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This is true, and in fact more has been shown in the recent preprint http://arxiv.org/abs/1005.3049 of Fang, Gao, and Smith. One can also give the following alternative argument based on ideas of Popa:

If $LH \subset LG$ is a MASA then it follows from the condition $ ( hgh^{-1} \ | \ h \in H ) = \infty$ for all $g \in G \setminus H$, that the normalizer of $H$ in $G$ is the same as the set of elements $g \in G$ such that $[H: H \cap gHg^{-1}] < \infty$. (This set is not in general closed under inversion but in this case it is since it coincides with the normalizer.)

Suppose we fix $g \in G$ such that $[H: H \cap gHg^{-1}] = \infty$ and let's show that $u_g$ is orthogonal to $\mathcal N_{LG}(LH)''$. Since $\mathcal N_{LG}(LH)''$ is spanned by $\mathcal N_{LG}(LH)$ it is enought to show that $u_g$ is orthogonal to this set and so let's fix $v \in \mathcal N_{LG}(LH)$.

Before we show that $u_g$ and $v$ are orthogonal let's rewrite the condition $[H: H \cap gHg^{-1}] = \infty$ in a more von Neumann algebraic friendly context which states that there are always "large" subalgebras of $LH$ which are almost moved orthogonal to $LH$. (Note that in the case that $H$ is free abelian the following lemma is trivial since subgroups then have complements.)

Lemma: For all $n \in \mathbb N, \delta > 0$ there exists a finite dimensional subalgebra $A_0 \subset LH$ such that if $p$ is any minimal projection in $A_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u_g^* p u_g - \tau(p) \rangle | < \delta \|x \|_2$ for all $x \in LH$.

Proof. This essentially follows from Popa's intertwining techniques since the condition $[H: H \cap gHg^{-1}] = \infty$ translates in this context to $LH \not\prec_{LH} L(H \cap gHg^{-1})$ (See Popa's paper http://www.ams.org/mathscinet-getitem?mr=2231961).

Let's show this by induction on $n$. For the case when $n = 1$ consider the group $\mathcal G = ( u \in \mathcal U(LH) \ | \ u = 1 - 2p, p \in \mathcal P(LH), \tau(p) = 1/2 ) \cup (1)$. Since $\mathcal G$ generates $LH$ as a von Neumann algebra and since $LH \not\prec_{LH} L(H \cap gHg^{-1})$ it follows from Popa's intertwining Theorem that there exists a sequence $p_k \in \mathcal P(LH)$ with $\tau(p_k) = 1/2$ such that $\lim_{k \to \infty} \| E_{L(H \cap gHg^{-1})}(1 - 2p_k ) \|_2 = 0$ (see Popa, op. cit.). In particular, for some $k$ this is less than $2\delta$ and so if $x \in LH$, $\| x \|_2 < 1$ we have $| \langle x, u_g^*p_ku_g - \tau(p) \rangle | \leq \| E_{LH}(u_g^* p u_g - \tau(p) ) \|$ $_2 = \| E_{L(H \cap gHg^{-1})} (p_k - 1/2) \|_2 < \delta$. The same inequality holds for the other minimal projection $1 - p_k$.

Once we have produced such an $A_0$ for $1/2^n$ then given any minimal projection $p \in A_0$ we again have that $pLH \not\prec_{pLH} pL(H \cap gHg^*)$ and so the argument above shows that there exists $p_1$ and $p_2$ in $\mathcal P(LH)$ such that $p_1 + p_2 = p$, each has half the trace and $| \langle x, u_g^* p_j u_g - \tau(p_j) \rangle | < \delta$. This proves the induction step. QED

Now that we have established the above lemma, the fact that $u_g$ and $v$ are orthogonal follows from a lemma of Popa's in http://www.ams.org/mathscinet-getitem?mr=703810. Let's give the proof here.

Let $\varepsilon > 0$ be given and take $n \in \mathbb N$ such that $1/2^n < \varepsilon/2$. From the above lemma let's consider a finite dimensional subalgebra $A_0 \subset LH$ such that if $p$ is any minimal projection in $A_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u_g^*pu_g - \tau(p) \rangle | < \| x \|_2 \varepsilon/2^{n + 1}$. Let's denote the minimal projections in $A_0$ by $p_k$ where $1 \leq k \leq 2^n$. Denote by $B_0$ the commutant of $A_0$ in $LG$.

Since $v \in \mathcal N_{LG}(LH)$ we have that $vLHv^* = LH$, hence $v^* p_k v \in LH$ for each $k$. Therefore $| \langle v, u_g \rangle |^2 \leq \| E_{B_0} ( vu_g^*) \|_2^2$ $= \| $ $\Sigma_k$ $ p_k v u_g^* p_k \|_2^2 = \Sigma_k \langle v^* p_k v, u_g^* p_k u_g \rangle < (\Sigma_k \tau(p_k)^2 ) + \Sigma_k \varepsilon/2^{n + 1} < \varepsilon$.

Since $\varepsilon$ was arbitrary we conclude that $u_g$ and $v$ are orthogonal. Hence since $v$ was arbitrary we conclude that $\mathcal N_{LG}(LH)'' = L(\mathcal N_G(H))$.

This is true, and in fact more has been shown in the recent preprint http://arxiv.org/abs/1005.3049 of Fang, Gao, and Smith. One can also give the following alternative argument based on ideas of Popa:

If $LH \subset LG$ is a MASA then it follows from the condition $ ( hgh^{-1} \ | \ h \in H ) = \infty$ for all $g \in G \setminus H$, that the normalizer of $H$ in $G$ is the same as the set of elements $g \in G$ such that $[H: H \cap gHg^{-1}] < \infty$. (This set is not in general closed under inversion but in this case it is since it coincides with the normalizer.)

Suppose we fix $g \in G$ such that $[H: H \cap gHg^{-1}] = \infty$ and let's show that $u_g$ is orthogonal to $\mathcal N_{LG}(LH)''$. Since $\mathcal N_{LG}(LH)''$ is spanned by $\mathcal N_{LG}(LH)$ it is enought to show that $u_g$ is orthogonal to this set and so let's fix $v \in \mathcal N_{LG}(LH)$.

Before we show that $u_g$ and $v$ are orthogonal let's rewrite the condition $[H: H \cap gHg^{-1}] = \infty$ in a more von Neumann algebraic friendly context which states that there are always "large" subalgebras of $LH$ which are almost moved orthogonal to $LH$.

Lemma: For all $n \in \mathbb N, \delta > 0$ there exists a finite dimensional subalgebra $A_0 \subset LH$ such that if $p$ is any minimal projection in $A_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u_g^* p u_g - \tau(p) \rangle | < \delta \|x \|_2$ for all $x \in LH$.

Proof. This essentially follows from Popa's intertwining techniques since the condition $[H: H \cap gHg^{-1}] = \infty$ translates in this context to $LH \not\prec_{LH} L(H \cap gHg^{-1})$ (See Popa's paper http://www.ams.org/mathscinet-getitem?mr=2231961).

Let's show this by induction on $n$. For the case when $n = 1$ consider the group $\mathcal G = ( u \in \mathcal U(LH) \ | \ u = 1 - 2p, p \in \mathcal P(LH), \tau(p) = 1/2 ) \cup (1)$. Since $\mathcal G$ generates $LH$ as a von Neumann algebra and since $LH \not\prec_{LH} L(H \cap gHg^{-1})$ it follows from Popa's intertwining Theorem that there exists a sequence $p_k \in \mathcal P(LH)$ with $\tau(p_k) = 1/2$ such that $\lim_{k \to \infty} \| E_{L(H \cap gHg^{-1})}(1 - 2p_k ) \|_2 = 0$ (see Popa, op. cit.). In particular, for some $k$ this is less than $2\delta$ and so if $x \in LH$, $\| x \|_2 < 1$ we have $| \langle x, u_g^*p_ku_g - \tau(p) \rangle | \leq \| E_{LH}(u_g^* p u_g - \tau(p) ) \|$ $_2 = \| E_{L(H \cap gHg^{-1})} (p_k - 1/2) \|_2 < \delta$. The same inequality holds for the other minimal projection $1 - p_k$.

Once we have produced such an $A_0$ for $1/2^n$ then given any minimal projection $p \in A_0$ we again have that $pLH \not\prec_{pLH} pL(H \cap gHg^*)$ and so the argument above shows that there exists $p_1$ and $p_2$ in $\mathcal P(LH)$ such that $p_1 + p_2 = p$, each has half the trace and $| \langle x, u_g^* p_j u_g - \tau(p_j) \rangle | < \delta$. This proves the induction step. QED

Now that we have established the above lemma, the fact that $u_g$ and $v$ are orthogonal follows from a lemma of Popa's in http://www.ams.org/mathscinet-getitem?mr=703810. Let's give the proof here.

Let $\varepsilon > 0$ be given and take $n \in \mathbb N$ such that $1/2^n < \varepsilon/2$. From the above lemma let's consider a finite dimensional subalgebra $A_0 \subset LH$ such that if $p$ is any minimal projection in $A_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u_g^*pu_g - \tau(p) \rangle | < \| x \|_2 \varepsilon/2^{n + 1}$. Let's denote the minimal projections in $A_0$ by $p_k$ where $1 \leq k \leq 2^n$. Denote by $B_0$ the commutant of $A_0$ in $LG$.

Since $v \in \mathcal N_{LG}(LH)$ we have that $vLHv^* = LH$, hence $v^* p_k v \in LH$ for each $k$. Therefore $| \langle v, u_g \rangle |^2 \leq \| E_{B_0} ( vu_g^*) \|_2^2$ $= \| $ $\Sigma_k$ $ p_k v u_g^* p_k \|_2^2 = \Sigma_k \langle v^* p_k v, u_g^* p_k u_g \rangle < (\Sigma_k \tau(p_k)^2 ) + \Sigma_k \varepsilon/2^{n + 1} < \varepsilon$.

Since $\varepsilon$ was arbitrary we conclude that $u_g$ and $v$ are orthogonal. Hence since $v$ was arbitrary we conclude that $\mathcal N_{LG}(LH)'' = L(\mathcal N_G(H))$.

This is true, and in fact more has been shown in the resent preprint http://arxiv.org/abs/1005.3049 of Fang, Gao, and Smith. One can also give the following alternative argument based on ideas of Popa:

If $LH \subset LG$ is a MASA then it follows from the condition $ ( hgh^{-1} \ | \ h \in H ) = \infty$ for all $g \in G \setminus H$, that the normalizer of $H$ in $G$ is the same as the set of elements $g \in G$ such that $[H: H \cap gHg^{-1}] < \infty$. (This set is not in general closed under inversion but in this case it is since it coincides with the normalizer.)

Suppose we fix $g \in G$ such that $[H: H \cap gHg^{-1}] = \infty$ and let's show that $u_g$ is orthogonal to $\mathcal N_{LG}(LH)''$. Since $\mathcal N_{LG}(LH)''$ is spanned by $\mathcal N_{LG}(LH)$ it is enought to show that $u_g$ is orthogonal to this set and so let's fix $v \in \mathcal N_{LG}(LH)$.

Before we show that $u_g$ and $v$ are orthogonal let's rewrite the condition $[H: H \cap gHg^{-1}] = \infty$ in a more von Neumann algebraic friendly context which states that there are always "large" subalgebras of $LH$ which are almost moved orthogonal to $LH$. (Note that in the case that $H$ is free abelian the following lemma is trivial since subgroups then have complements.)

Lemma: For all $n \in \mathbb N, \delta > 0$ there exists a finite dimensional subalgebra $A_0 \subset LH$ such that if $p$ is any minimal projection in $A_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u_g^* p u_g - \tau(p) \rangle | < \delta \|x \|_2$ for all $x \in LH$.

Proof. This essentially follows from Popa's intertwining techniques since the condition $[H: H \cap gHg^{-1}] = \infty$ translates in this context to $LH \not\prec_{LH} L(H \cap gHg^{-1})$ (See Popa's paper http://www.ams.org/mathscinet-getitem?mr=2231961).

Let's show this by induction on $n$. For the case when $n = 1$ consider the group $\mathcal G = ( u \in \mathcal U(LH) \ | \ u = 1 - 2p, p \in \mathcal P(LH), \tau(p) = 1/2 ) \cup (1)$. Since $\mathcal G$ generates $LH$ as a von Neumann algebra and since $LH \not\prec_{LH} L(H \cap gHg^{-1})$ it follows from Popa's intertwining Theorem that there exists a sequence $p_k \in \mathcal P(LH)$ with $\tau(p_k) = 1/2$ such that $\lim_{k \to \infty} \| E_{L(H \cap gHg^{-1})}(1 - 2p_k ) \|_2 = 0$ for all $x \in LH$ (see Popa, op. cit.). In particular, for some $k$ this is less than $2\delta$ and so if $x \in LH$, $\| x \|_2 < 1$ we have $| \langle x, u_g^*p_ku_g - \tau(p) \rangle | \leq \| E_{LH}(u_g^* p u_g - \tau(p) ) \|$ $_2 = \| E_{L(H \cap gHg^{-1})} (p_k - 1/2) \|_2 < \delta$. The same inequality holds for the other minimal projection $1 - p_k$.

Once we have produced such an $A_0$ for $1/2^n$ then given any minimal projection $p \in A_0$ we again have that $pLH \not\prec_{pLH} pL(H \cap gHg^*)$ and so the argument above shows that there exists $p_1$ and $p_2$ in $\mathcal P(LH)$ such that $p_1 + p_2 = p$, each has half the trace and $| \langle x, u_g^* p_j u_g - \tau(p_j) \rangle | < \delta$. This proves the induction step. QED

Now that we have established the above lemma, the fact that $u_g$ and $v$ are orthogonal follows from a lemma of Popa's in http://www.ams.org/mathscinet-getitem?mr=703810. Let's give the proof here.

Let $\varepsilon > 0$ be given and take $n \in \mathbb N$ such that $1/2^n < \varepsilon/2$. From the above lemma let's consider a finite dimensional subalgebra $A_0 \subset LH$ such that if $p$ is any minimal projection in $A_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u_g^*pu_g - \tau(p) \rangle | < \| x \|_2 \varepsilon/2^{n + 1}$. Let's denote the minimal projections in $A_0$ by $p_k$ where $1 \leq k \leq 2^n$. Denote by $B_0$ the commutant of $A_0$ in $LG$.

Since $v \in \mathcal N_{LG}(LH)$ we have that $vLHv^* = LH$, hence $v^* p_k v \in LH$ for each $k$. Therefore $| \langle v, u_g \rangle |^2 \leq \| E_{B_0} ( vu_g^*) \|_2^2$ $= \| $ $\Sigma_k$ $ p_k v u_g^* p_k \|_2^2 = \Sigma_k \langle v^* p_k v, u_g^* p_k u_g \rangle < (\Sigma_k \tau(p_k)^2 ) + \Sigma_k \varepsilon/2^{n + 1} < \varepsilon$.

Since $\varepsilon$ was arbitrary we conclude that $u_g$ and $v$ are orthogonal. Hence since $v$ was arbitrary we conclude that $\mathcal N_{LG}(LH)'' = L(\mathcal N_G(H))$.

This is true, and in fact more has been shown in the recent preprint http://arxiv.org/abs/1005.3049 of Fang, Gao, and Smith. One can also give the following alternative argument based on ideas of Popa:

If $LH \subset LG$ is a MASA then it follows from the condition $ ( hgh^{-1} \ | \ h \in H ) = \infty$ for all $g \in G \setminus H$, that the normalizer of $H$ in $G$ is the same as the set of elements $g \in G$ such that $[H: H \cap gHg^{-1}] < \infty$. (This set is not in general closed under inversion but in this case it is since it coincides with the normalizer.)

Suppose we fix $g \in G$ such that $[H: H \cap gHg^{-1}] = \infty$ and let's show that $u_g$ is orthogonal to $\mathcal N_{LG}(LH)''$. Since $\mathcal N_{LG}(LH)''$ is spanned by $\mathcal N_{LG}(LH)$ it is enought to show that $u_g$ is orthogonal to this set and so let's fix $v \in \mathcal N_{LG}(LH)$.

Before we show that $u_g$ and $v$ are orthogonal let's rewrite the condition $[H: H \cap gHg^{-1}] = \infty$ in a more von Neumann algebraic friendly context which states that there are always "large" subalgebras of $LH$ which are almost moved orthogonal to $LH$. (Note that in the case that $H$ is free abelian the following lemma is trivial since subgroups then have complements.)

Lemma: For all $n \in \mathbb N, \delta > 0$ there exists a finite dimensional subalgebra $A_0 \subset LH$ such that if $p$ is any minimal projection in $A_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u_g^* p u_g - \tau(p) \rangle | < \delta \|x \|_2$ for all $x \in LH$.

Proof. This essentially follows from Popa's intertwining techniques since the condition $[H: H \cap gHg^{-1}] = \infty$ translates in this context to $LH \not\prec_{LH} L(H \cap gHg^{-1})$ (See Popa's paper http://www.ams.org/mathscinet-getitem?mr=2231961).

Let's show this by induction on $n$. For the case when $n = 1$ consider the group $\mathcal G = ( u \in \mathcal U(LH) \ | \ u = 1 - 2p, p \in \mathcal P(LH), \tau(p) = 1/2 ) \cup (1)$. Since $\mathcal G$ generates $LH$ as a von Neumann algebra and since $LH \not\prec_{LH} L(H \cap gHg^{-1})$ it follows from Popa's intertwining Theorem that there exists a sequence $p_k \in \mathcal P(LH)$ with $\tau(p_k) = 1/2$ such that $\lim_{k \to \infty} \| E_{L(H \cap gHg^{-1})}(1 - 2p_k ) \|_2 = 0$ (see Popa, op. cit.). In particular, for some $k$ this is less than $2\delta$ and so if $x \in LH$, $\| x \|_2 < 1$ we have $| \langle x, u_g^*p_ku_g - \tau(p) \rangle | \leq \| E_{LH}(u_g^* p u_g - \tau(p) ) \|$ $_2 = \| E_{L(H \cap gHg^{-1})} (p_k - 1/2) \|_2 < \delta$. The same inequality holds for the other minimal projection $1 - p_k$.

Once we have produced such an $A_0$ for $1/2^n$ then given any minimal projection $p \in A_0$ we again have that $pLH \not\prec_{pLH} pL(H \cap gHg^*)$ and so the argument above shows that there exists $p_1$ and $p_2$ in $\mathcal P(LH)$ such that $p_1 + p_2 = p$, each has half the trace and $| \langle x, u_g^* p_j u_g - \tau(p_j) \rangle | < \delta$. This proves the induction step. QED

Now that we have established the above lemma, the fact that $u_g$ and $v$ are orthogonal follows from a lemma of Popa's in http://www.ams.org/mathscinet-getitem?mr=703810. Let's give the proof here.

Let $\varepsilon > 0$ be given and take $n \in \mathbb N$ such that $1/2^n < \varepsilon/2$. From the above lemma let's consider a finite dimensional subalgebra $A_0 \subset LH$ such that if $p$ is any minimal projection in $A_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u_g^*pu_g - \tau(p) \rangle | < \| x \|_2 \varepsilon/2^{n + 1}$. Let's denote the minimal projections in $A_0$ by $p_k$ where $1 \leq k \leq 2^n$. Denote by $B_0$ the commutant of $A_0$ in $LG$.

Since $v \in \mathcal N_{LG}(LH)$ we have that $vLHv^* = LH$, hence $v^* p_k v \in LH$ for each $k$. Therefore $| \langle v, u_g \rangle |^2 \leq \| E_{B_0} ( vu_g^*) \|_2^2$ $= \| $ $\Sigma_k$ $ p_k v u_g^* p_k \|_2^2 = \Sigma_k \langle v^* p_k v, u_g^* p_k u_g \rangle < (\Sigma_k \tau(p_k)^2 ) + \Sigma_k \varepsilon/2^{n + 1} < \varepsilon$.

Since $\varepsilon$ was arbitrary we conclude that $u_g$ and $v$ are orthogonal. Hence since $v$ was arbitrary we conclude that $\mathcal N_{LG}(LH)'' = L(\mathcal N_G(H))$.

This is true, and in fact more has been shown in the resent preprint http://arxiv.org/abs/1005.3049 of Fang, Gao, and Smith. One can also give the following alternative argument based on ideas of Popa:

If $LH \subset LG$ is a MASA then it follows from the condition $ ( hgh^{-1} \ | \ h \in H ) = \infty$ for all $g \in G \setminus H$, that the normalizer of $H$ in $G$ is the same as the set of elements $g \in G$ such that $[H: H \cap gHg^{-1}] < \infty$. (This set is not in general closed under inversion but in this case it is since it coincides with the normalizer.)

Suppose we fix $g \in G$ such that $[H: H \cap gHg^{-1}] = \infty$ and let's show that $u_g$ is orthogonal to $\mathcal N_{LG}(LH)''$. Since $\mathcal N_{LG}(LH)''$ is spanned by $\mathcal N_{LG}(LH)$ it is enought to show that $u_g$ is orthogonal to this set and so let's fix $v \in \mathcal N_{LG}(LH)$.

Before we show that $u_g$ and $v$ are orthogonal let's rewrite the condition $[H: H \cap gHg^{-1}] = \infty$ in a more von Neumann algebraic friendly context which states that there are always "large" subalgebras of $LH$ which are almost moved orthogonal to $LH$. (Note that in the case that $H$ is free abelian the following lemma is trivial since subgroups then have complements.)

Lemma: For all $n \in \mathbb N, \delta > 0$ there exists a finite dimensional subalgebra $A_0 \subset LH$ such that if $p$ is any minimal projection in $A_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u_g^* p u_g - \tau(p) \rangle | < \delta \|x \|_2$ for all $x \in LH$.

Proof. This essentially follows from Popa's intertwining techniques since the condition $[H: H \cap gHg^{-1}] = \infty$ translates in this context to $LH \not\prec_{LH} L(H \cap gHg^{-1})$ (See Popa's paper http://www.ams.org/mathscinet-getitem?mr=2231961).

Let's show this by induction on $n$. For the case when $n = 1$ consider the group $\mathcal G = ( u \in \mathcal U(LH) \ | \ u = 1 - 2p, p \in \mathcal P(LH), \tau(p) = 1/2 ) \cup (1)$. Since $\mathcal G$ generates $LH$ as a von Neumann algebra and since $LH \not\prec_{LH} L(H \cap gHg^{-1})$ it follows from Popa's intertwining Theorem that there exists a sequence $p_k \in \mathcal P(LH)$ with $\tau(p_k) = 1/2$ such that $\lim_{k \to \infty} \| E_{L(H \cap gHg^{-1})}(1 - 2p_k ) \|_2 = 0$ for all $x \in LH$ (see Popa, op. cit.). In particular, for some $k$ this is less than $2\delta$ and so if $x \in LH$, $\| x \|_2 < 1$ we have $| \langle x, u_g^*p_ku_g - \tau(p) \rangle |$ $\leq \| E_{LH}(u_g^* p u_g - \tau(p) )$ $ \| $ $_2$ $= \| E_{L(H \cap gHg^{-1})} (p_k - 1/2) \|_2 < \delta$. The same inequality holds for the other minimal projection $1 - p_k$.

Once we have produced such an $A_0$ for $1/2^n$ then given any minimal projection $p \in A_0$ we again have that $pLH \not\prec_{pLH} pL(H \cap gHg^*)$ and so the argument above shows that there exists $p_1$ and $p_2$ in $\mathcal P(LH)$ such that $p_1 + p_2 = p$, each has half the trace and $| \langle x, u_g^* p_j u_g - \tau(p_j) \rangle | < \delta$. This proves the induction step. QED

Now that we have established the above lemma, the fact that $u_g$ and $v$ are orthogonal follows from a lemma of Popa's in http://www.ams.org/mathscinet-getitem?mr=703810. Let's give the proof here.

Let $\varepsilon > 0$ be given and take $n \in \mathbb N$ such that $1/2^n < \varepsilon/2$. From the above lemma let's consider a finite dimensional subalgebra $A_0 \subset LH$ such that if $p$ is any minimal projection in $A_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u_g^*pu_g - \tau(p) \rangle | < \| x \|_2 \varepsilon/2^{n + 1}$. Let's denote the minimal projections in $A_0$ by $p_k$ where $1 \leq k \leq 2^n$. Denote by $B_0$ the commutant of $A_0$ in $LG$.

Since $v \in \mathcal N_{LG}(LH)$ we have that $vLHv^* = LH$, hence $v^* p_k v \in LH$ for each $k$. Therefore $| \langle v, u_g \rangle |^2 \leq \| E_{B_0} ( vu_g^*) \|_2^2$ $= \| $ $\Sigma_k$ $ p_k v u_g^* p_k \|_2^2 = \Sigma_k \langle v^* p_k v, u_g^* p_k u_g \rangle < (\Sigma_k \tau(p_k)^2 ) + \Sigma_k \varepsilon/2^{n + 1} < \varepsilon$.

Since $\varepsilon$ was arbitrary we conclude that $u_g$ and $v$ are orthogonal. Hence since $v$ was arbitrary we conclude that $\mathcal N_{LG}(LH)'' = L(\mathcal N_G(H))$.

This is true, and in fact more has been shown in the resent preprint http://arxiv.org/abs/1005.3049 of Fang, Gao, and Smith. One can also give the following alternative argument based on ideas of Popa:

If $LH \subset LG$ is a MASA then it follows from the condition $ ( hgh^{-1} \ | \ h \in H ) = \infty$ for all $g \in G \setminus H$, that the normalizer of $H$ in $G$ is the same as the set of elements $g \in G$ such that $[H: H \cap gHg^{-1}] < \infty$. (This set is not in general closed under inversion but in this case it is since it coincides with the normalizer.)

Suppose we fix $g \in G$ such that $[H: H \cap gHg^{-1}] = \infty$ and let's show that $u_g$ is orthogonal to $\mathcal N_{LG}(LH)''$. Since $\mathcal N_{LG}(LH)''$ is spanned by $\mathcal N_{LG}(LH)$ it is enought to show that $u_g$ is orthogonal to this set and so let's fix $v \in \mathcal N_{LG}(LH)$.

Before we show that $u_g$ and $v$ are orthogonal let's rewrite the condition $[H: H \cap gHg^{-1}] = \infty$ in a more von Neumann algebraic friendly context which states that there are always "large" subalgebras of $LH$ which are almost moved orthogonal to $LH$. (Note that in the case that $H$ is free abelian the following lemma is trivial since subgroups then have complements.)

Lemma: For all $n \in \mathbb N, \delta > 0$ there exists a finite dimensional subalgebra $A_0 \subset LH$ such that if $p$ is any minimal projection in $A_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u_g^* p u_g - \tau(p) \rangle | < \delta \|x \|_2$ for all $x \in LH$.

Proof. This essentially follows from Popa's intertwining techniques since the condition $[H: H \cap gHg^{-1}] = \infty$ translates in this context to $LH \not\prec_{LH} L(H \cap gHg^{-1})$ (See Popa's paper http://www.ams.org/mathscinet-getitem?mr=2231961).

Let's show this by induction on $n$. For the case when $n = 1$ consider the group $\mathcal G = ( u \in \mathcal U(LH) \ | \ u = 1 - 2p, p \in \mathcal P(LH), \tau(p) = 1/2 ) \cup (1)$. Since $\mathcal G$ generates $LH$ as a von Neumann algebra and since $LH \not\prec_{LH} L(H \cap gHg^{-1})$ it follows from Popa's intertwining Theorem that there exists a sequence $p_k \in \mathcal P(LH)$ with $\tau(p_k) = 1/2$ such that $\lim_{k \to \infty} \| E_{L(H \cap gHg^{-1})}(1 - 2p_k ) \|_2 = 0$ for all $x \in LH$ (see Popa, op. cit.). In particular, for some $k$ this is less than $2\delta$ and so if $x \in LH$, $\| x \|_2 < 1$ we have $| \langle x, u_g^*p_ku_g - \tau(p) \rangle | \leq \| E_{LH}(u_g^* p u_g - \tau(p) ) \|$ $_2 = \| E_{L(H \cap gHg^{-1})} (p_k - 1/2) \|_2 < \delta$. The same inequality holds for the other minimal projection $1 - p_k$.

Once we have produced such an $A_0$ for $1/2^n$ then given any minimal projection $p \in A_0$ we again have that $pLH \not\prec_{pLH} pL(H \cap gHg^*)$ and so the argument above shows that there exists $p_1$ and $p_2$ in $\mathcal P(LH)$ such that $p_1 + p_2 = p$, each has half the trace and $| \langle x, u_g^* p_j u_g - \tau(p_j) \rangle | < \delta$. This proves the induction step. QED

Now that we have established the above lemma, the fact that $u_g$ and $v$ are orthogonal follows from a lemma of Popa's in http://www.ams.org/mathscinet-getitem?mr=703810. Let's give the proof here.

Let $\varepsilon > 0$ be given and take $n \in \mathbb N$ such that $1/2^n < \varepsilon/2$. From the above lemma let's consider a finite dimensional subalgebra $A_0 \subset LH$ such that if $p$ is any minimal projection in $A_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u_g^*pu_g - \tau(p) \rangle | < \| x \|_2 \varepsilon/2^{n + 1}$. Let's denote the minimal projections in $A_0$ by $p_k$ where $1 \leq k \leq 2^n$. Denote by $B_0$ the commutant of $A_0$ in $LG$.

Since $v \in \mathcal N_{LG}(LH)$ we have that $vLHv^* = LH$, hence $v^* p_k v \in LH$ for each $k$. Therefore $| \langle v, u_g \rangle |^2 \leq \| E_{B_0} ( vu_g^*) \|_2^2$ $= \| $ $\Sigma_k$ $ p_k v u_g^* p_k \|_2^2 = \Sigma_k \langle v^* p_k v, u_g^* p_k u_g \rangle < (\Sigma_k \tau(p_k)^2 ) + \Sigma_k \varepsilon/2^{n + 1} < \varepsilon$.

Since $\varepsilon$ was arbitrary we conclude that $u_g$ and $v$ are orthogonal. Hence since $v$ was arbitrary we conclude that $\mathcal N_{LG}(LH)'' = L(\mathcal N_G(H))$.