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Let $G$ be a finite group such that $|\operatorname{Inn}(G)|=6$ and $Z(G)$ is an elementary abelian $2$-group of order $2^n$. Then prove or disprove that $G$ is isomorphic to $$\underbrace{C_{2}\times...\times C_{2}}_{n-1}\times D_{12}\hspace{10pt} \mathrm{or}\hspace{10pt} \underbrace{C_{2}\times...\times C_{2}}_{n-1}\times T,$$ where $D_{12}$ is the dihedral group of order 12 and $T=C_{3}\rtimes C_{4}$.

Attempt: This is true for groups of order $12$, $24$, $48$ or $96$. I think this is true in general.

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Yes I think it is true. Here are some suggestions. Prove that $G$ has a normal Sylow 3-subgroup and an abelian Sylow 2-subgroup, which must be $C_2^{n+1}$ or $C_2^n \times C_4$, giving the two cases. Note also that $D_{12} \cong D_6 \times C_2$. –  Derek Holt Jan 14 '13 at 19:46
@ Derek Holt: your hint is very nice. I think since $S_{3}$ has only one subgroup of order 3. Hence, we can conclude that Sylow 3-subgroup of $G$ is normal. I do not know why Sylow 2-subgroup of $G$ is abelian? But if it is abelian then it must be $C_{2}^{n+1}$ or $C_{2}^{n-1}\times C_{4}$, since $exp(G)$ is at most 12. Now by Schur-Zassenhaus Theorem, we have $$C_{3}\rtimes C_{2}^{n+1}\hspace{10pt} or \hspace{10pt} C_{3}\rtimes (C_{4}\times C_{2}^{n-1}).$$ Now do we can say these are the result? –  maryam Jan 15 '13 at 8:43
The Sylow 2-subgroup $S$ is abelian, because $S/Z(S)$ has order 2 and hence is cyclic. For the semidirect products, remember that the subgroup $Z(G)$ of index 2 in $S$ centralizes $C_3$, and elements in $S \setminus Z(G)$ invert the generator of $T$, so in fact the actions in the semidirect products are uniquely determined, and hence there is a unique group in both cases. –  Derek Holt Jan 15 '13 at 9:53
In this case yes, because we know that the subgroup $C_2^{n-1}$ is in the centre. –  Derek Holt Jan 15 '13 at 12:00
@ Derek Holt: if $H, K, M$ be subgroups of a group $G$ such that $M\subseteq Z(G)$, then do we can say $H\rtimes (K\times M)\cong(H\rtimes K)\times M$? –  maryam Jan 15 '13 at 13:16

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