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Let $P$ be an infinite extra special $p$-group for some prime $p$, namely, $Z(P)=P'=\Phi(P)$ and $P/Z(P)$ is infinite elementary abelian.

Let $C$ be a Prufer $q$-group for some prime $q\neq p$.

Question Is it possible to find an action of $C$ on $P$ such that $C$ acts irreducibly on $P/Z(P)$? (i.e. $P/Z(P)$ does not contains any $C$-invariant subgroup)

Unfortunately I cannot find any information about the automorphism group of an extraspecial group in the infinite case. Probably, it is just some kind of infinite symplectic group but I cannot figure out if such an action is possible or not.

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    $\begingroup$ Sorry I haven't time to work this out in detail right now, so I will comment rather than answer, but I think the answer is yes, but we should probably assume that $P$ has exponent $p$ to get the full symplectic group acting. Suppose for example that $p=5$ and $q=3$. We can find $g_1$ of order $3$ acting irreducibly in ${\rm Sp}(2,5)$. Then $g_2$ of order $9$ acting irreducible in ${\rm Sp}(6,5)$ with $g_2^3$ acting as three copies of $g_1$. Continue, choosing $g_n$ acting irreducibly in ${\rm Sp}(2.3^{n-1}.5)$ with $g_n^3$ acting as three copies of $g_{n-1}$. $\endgroup$ – Derek Holt Feb 28 '17 at 22:58
  • $\begingroup$ @DerekHolt Is there any chance you can write some more details? Unfortunately I do not know a lot about symplectic groups $\endgroup$ – W4cc0 Mar 1 '17 at 10:46
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My construction does not work for all $p$ and $q$. I need to assume that $p$ and $q$ are both odd, that $P$ is of exponent $p$, and that the multiplicative order of $p$ modulo $q$ is even. I don't know whether such an action exists for other $p,q$ - I would guess not.

Let us assume that $P$ is the central product of extraspecial groups $\langle x_i,y_i \rangle$ of order $p^3$ and exponent $p$ for $i \ge 1$. Note that the automorphism group of a central product of $n$ such groups is ${\rm Sp}(2n,p)$.

Now the group ${\rm Sp}(2n,p)$ has elements of order $p^n+1$. (To see that you observe that ${\rm Sp}(2,p^n) = {\rm SL}(2,p^n)$ has such elements and ${\rm Sp}(2,p^n)$ is naturally a subgroup of ${\rm Sp}(2n,p)$.)

Suppose that $2k$ is the order of $p$ modulo $q$, so $q|p^k+1$. Let $t \ge 1$ be maximal such that $q^t|p^k+1$. Then it is an easy exercise to show that the order of $p$ modulo $q^{t+i}$ for any $i \ge 0$ is $2kq^i$.

So, for a given $i > 0$, $q^{t+i}|p^{kq^i}+1$, and so ${\rm Sp}(2kq^i,p)$ has elements $g$ of order $q^{t+i}$. Such elements must act irreducibly on their natural module $V$. (All faithful irreducible representations of a cyclic group $C_m$ over a finite field of order $q$ have dimension the order of $q$ modulo $m$.) Then $g^q$ has order $q^{t+i-1}$ and, under the action of $g^q$, $V$ must decompose into a direct sum of $q$ (isomorphic) irreducible modules $V_i$.

Now the restriction of the symplectic form to each $V_i$ must be non-degenerate. The only other possibility would be that this restriction was totally singular, but that cannot happen with $q$ odd. (I could try and give more details about that if you wanted.)

Now we can put all of these ingredients together to construct an irreducible action of the Prufer $q$-group $C$ on $P$.

Let $g_0 \in C$ have order $q^t$. Then we can define an action of $g_0$ on $P$ by letting it act faithfully and irreducibly on the subgroups $\langle x_i,y_i : 1 \le i \le k \rangle$, $\langle x_i,y_i : k+1 \le i \le 2k \rangle$, etc.

Then, if $g_1$ is an element of $C$ of order $q^{t+1}$ with $g_1^q = g_0$, we can define the action of $g_1$ so that it acts irreducibly on the subgroups $\langle x_i,y_i : 1 \le i \le kq \rangle$, $\langle x_i,y_i: kq+1 \le i \le 2kq \rangle$, etc, and such that the action of $g_1^q$ is the same as that of $g_0$.

Now you can continue this construction by defining actions of $g_i$ consistent with those of $g_{i-1}$ for all $i > 0$. It is easy to see that the resulting action of $C$ on $P$ is irreducible.

Later addition: For the non-degeneracy of the $V_i$, if we could find some non-degenerate invaraint subspace $V_1$ for $g^q$, then their images under $g$ would give the required decomposition $V = V_1 \oplus \cdots \oplus V_q$ with each $V_i$ non-degenerate.

So let us prove by induction on $n$ that any element $x$ of order coprime to $p$ in ${\rm Sp}(2n,p)$ for which irreducible modules have dimension $m$ with $2n/m$ odd must stabilize and act irreducibly on some non-degenerate subspace. If not, then $x$ stabilizes and acts irreducibly on a totally singular subspace $W$ of dimension $m$. Then the subspace $W^\perp$ of $V$ that annihilates $W$ under the symplectic form contains $W$ and has dimension $2n-m$, and the form induced on $W^\perp/W$ is non-degenerate.

Now $x$ fixes the subspace $W^\perp$, so it fixes a complement $X$ of $W$ in $W^\perp$ of dimension $2n-2m$, and the symplectic form induced on $X$ is non-degenerate, so our claim follows by applying induction to the action on $X$, which is an element of ${\rm Sp}(2(n-m)p)$.

That argument seems a bit awkward and their may be a better way of seeing this, but I think it works.

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  • $\begingroup$ Ok, maybe some more details on why $V_i$ is non-degenerate could be useful for me. $\endgroup$ – W4cc0 Mar 2 '17 at 10:28

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