True or false: If G is special 2-group, then |G:Z(G)| is a square.
Recall that G is a special p-group if Z(G) = G' = Frat (G) be elementary abelian. The above assertion is true, whenever |Z(G)| = 2, that is, G is an extra-special 2-group. Thank you.
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3 Answers
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Being systematic a $2$-special group is specified completely by two $\mathbb Z/2$-vector spaces $U=G/Z(G)$ and $V=Z(G)$ together with the induced square map $\Gamma^2U\to V$ which is non-degenerate in the following sense: We have an injective map $\Lambda^2U\to\Gamma^2U$ mapping $u\land v\mapsto\gamma_1(u)\gamma_1(v)$ so that the composite $\Lambda^2U\to\Gamma^2U\to V$ is non-degenerate. Every map $\Gamma^2U\to V$ occurs for some central extension of $V$ by $U$ and the extension gives a $2$-special group precisely when $\Lambda^2U\to\Gamma^2U\to V$ is non-degenerate.
For $V=(\mathbb Z/2)^2$ a map $\Lambda^2U\to V$ corresponds to a pair of alternating forms on $U$ and the non-degeneracy says exactly that the radicals of the two forms intersect in zero. Now, if $V$ is even-dimensional there is a single non-degenerate alternating form and if $V$ is odd-dimensional, then any $1$-dimensional subspace is the radical of some alternating form so that if $\dim V>1$ there are always two forms whose radicals intersect trivially. Picking such a map $\Lambda^2U\to V$ we simply may take any linear extension to a map $\Gamma^2U\to V$ to get a $2$-special group. This means that the answer to the question is no.
answered Mar 14, 2011 at 14:48
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This is false. Let $G = E_{16} \rtimes C_2$, where $E_{16}$ denotes the elementary abelian group of order $16$. Then $G$ is special, but $Z(G) \cong C_2 \times C_2$ has index $8$ in $G$.
answered Mar 14, 2011 at 14:25
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3$\begingroup$ To define $G$ unambiguously you need to specify the action of the $C_2$ on the $E_{16}$. An example of such an action that results in a special group $G$ is $a \rightarrow a$, $b \rightarrow b$, $c \rightarrow ca$, $d \rightarrow db$, where $a,b,c,d$ are generators of $E_{16}$. $\endgroup$ Commented Mar 14, 2011 at 16:49
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$\begingroup$ @Derek: I fully agree, sorry for this omission. $\endgroup$ Commented Mar 15, 2011 at 7:18
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The Sylow 2-subgroups of the Suzuki groups furnish an infinite sequence of special 2-groups that fail to satisfy this property.