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Now let $G=G(k)$ be the free group of rank $k$ in this variety. Clearly $G/G^2$ is isomorphic to $Q_k$, so it remains to determine the multiplicities $n_\chi=n_\chi(k)$. The abelianization of $G(k)$ being isomorphic to $C_6^k$, we have $n_0(k)=k$. We have a canonical homomorphism $\mathrm{Aut}(G)\to\mathrm{Aut}(Q_k)\simeq\mathrm{GL}_k(\mathbf{F}_2)$. It is surjective: indeed if $(x_1,\dots,x_k)$ is a basis of $G$, then mapping $x_i$ to $x_ix_j$ ($i\neq j$ fixed) and fixing all other basis elements, induces the corresponding elementary matrix, and these generate $\mathrm{GL}_k(\mathbf{F}_2)=\mathrm{SL}_k(\mathbf{F}_2)$. It follows that all $n_\chi(k)$, for $\chi\neq 0$, are equal, say to some number $n(k)$, to determine.

(2) $V_0^k\rtimes Q_k$ has the same property: this is obvious since it is isomorphic to $C_6^k$.

Now let $G=G(k)$ be the free group of rank $k$ in this variety. Clearly $G/G^2$ is isomorphic to $Q_k$, so it remains to determine the multiplicities $n_\chi=n_\chi(k)$. The abelianization of $G(k)$ being isomorphic to $C_{2p}^k$, we have $n_0(k)=k$. We have a canonical homomorphism $\mathrm{Aut}(G)\to\mathrm{Aut}(Q_k)\simeq\mathrm{GL}_k(\mathbf{F}_2)$. It is surjective: indeed if $(x_1,\dots,x_k)$ is a basis of $G$, then mapping $x_i$ to $x_ix_j$ ($i\neq j$ fixed) and fixing all other basis elements, induces the corresponding elementary matrix, and these generate $\mathrm{GL}_k(\mathbf{F}_2)=\mathrm{SL}_k(\mathbf{F}_2)$. It follows that all $n_\chi(k)$, for $\chi\neq 0$, are equal, say to some number $n(k)$, to determine.

(2) $V_0^k\rtimes Q_k$ has the same property: this is obvious since it is isomorphic to $C_{2p}^k$.

Now let $G=G(k)$ be the free group of rank $k$ in this variety. Clearly $G/G^2$ is isomorphic to $Q_k$, so it remains to determine the multiplicities $n_\chi=n_\chi(k)$. The abelianization of $G(k)$ being isomorphic to $C_2^k$, we have $n_0(k)=k$. We have a canonical homomorphism $\mathrm{Aut}(G)\to\mathrm{Aut}(Q_k)\simeq\mathrm{GL}_k(\mathbf{F}_2)$. It is surjective: indeed if $(x_1,\dots,x_k)$ is a basis of $G$, then mapping $x_i$ to $x_ix_j$ ($i\neq j$ fixed) and fixing all other basis elements, induces the corresponding elementary matrix, and these generate $\mathrm{GL}_k(\mathbf{F}_2)=\mathrm{SL}_k(\mathbf{F}_2)$. It follows that all $n_\chi(k)$, for $\chi\neq 0$, are equal, say to some number $n(k)$, to determine.

Now let $G=G(k)$ be the free group of rank $k$ in this variety. Clearly $G/G^2$ is isomorphic to $Q_k$, so it remains to determine the multiplicities $n_\chi=n_\chi(k)$. The abelianization of $G(k)$ being isomorphic to $C_6^k$, we have $n_0(k)=k$. We have a canonical homomorphism $\mathrm{Aut}(G)\to\mathrm{Aut}(Q_k)\simeq\mathrm{GL}_k(\mathbf{F}_2)$. It is surjective: indeed if $(x_1,\dots,x_k)$ is a basis of $G$, then mapping $x_i$ to $x_ix_j$ ($i\neq j$ fixed) and fixing all other basis elements, induces the corresponding elementary matrix, and these generate $\mathrm{GL}_k(\mathbf{F}_2)=\mathrm{SL}_k(\mathbf{F}_2)$. It follows that all $n_\chi(k)$, for $\chi\neq 0$, are equal, say to some number $n(k)$, to determine.

First observe that $D_{2p}$ satisfies the group identities $x^{2p}=1$, $[x^2,y^2]$.

In addition, the identity $[x^2,y^2]$ implies that for $G\in G_p$, any product of squares belongs to $G_p$, so the subgroup generated by $G^2(=G_p)$ is contained in $G_p$. This means that $G_p$ is a subgroup, obviously a normal elementary abelian $p$-subgroup. Since $G_p=G^2$, $G/G^2$ is a 2-group. If $G$ is finite, we deduce that $G=G_p\rtimes Q$, where $Q$ is any 2-Sylow subgroup, and $Q$ is elementary abelian (say of order $2^k$); for convenience write $Q=Q_k$.

Now let $G(k)$ be the free group of rank $k$ in this variety. Clearly $G/G^2$ is isomorphic to $Q_k$, so it remains to determine the multiplicities $n_\chi=n_\chi(k)$. The abelianization of $G(k)$ being isomorphic to $C_6^k$, we have $n_0(k)=k$. We have a canonical homomorphism $\mathrm{Aut}(G)\to\mathrm{Aut}(Q_k)\simeq\mathrm{GL}_2(\mathbf{F}_2)$. It is surjective: indeed if $(x_1,\dots,x_k)$ is a basis of $G$, then mapping $x_i$ to $x_ix_j$ ($i\neq j$ fixed) and fixing all other basis elements, induces the corresponding elementary matrix, and these generate $\mathrm{GL}_2(\mathbf{F}_2)=\mathrm{SL}_2(\mathbf{F}_2)$. It follows that all $n_\chi(k)$, for $\chi\neq 0$, are equal, say to some number $n(k)$, to determine.

First observe that $D_{2p}$ satisfies the group identities $x^{2p}=1$, $[x^2,y^2]=1$.

In addition, the identity $[x^2,y^2]=1$ implies that for $G\in G_p$, any product of squares belongs to $G_p$, so the subgroup generated by $G^2(=G_p)$ is contained in $G_p$. This means that $G_p$ is a subgroup, obviously a normal elementary abelian $p$-subgroup. Since $G_p=G^2$, $G/G^2$ is a 2-group. If $G$ is finite, we deduce that $G=G_p\rtimes Q$, where $Q$ is any 2-Sylow subgroup, and $Q$ is elementary abelian (say of order $2^k$); for convenience write $Q=Q_k$.

Now let $G=G(k)$ be the free group of rank $k$ in this variety. Clearly $G/G^2$ is isomorphic to $Q_k$, so it remains to determine the multiplicities $n_\chi=n_\chi(k)$. The abelianization of $G(k)$ being isomorphic to $C_2^k$, we have $n_0(k)=k$. We have a canonical homomorphism $\mathrm{Aut}(G)\to\mathrm{Aut}(Q_k)\simeq\mathrm{GL}_k(\mathbf{F}_2)$. It is surjective: indeed if $(x_1,\dots,x_k)$ is a basis of $G$, then mapping $x_i$ to $x_ix_j$ ($i\neq j$ fixed) and fixing all other basis elements, induces the corresponding elementary matrix, and these generate $\mathrm{GL}_k(\mathbf{F}_2)=\mathrm{SL}_k(\mathbf{F}_2)$. It follows that all $n_\chi(k)$, for $\chi\neq 0$, are equal, say to some number $n(k)$, to determine.