Order of automorphism group of an almost elementary abelian group

Question

Let $A$ be an almost elementary abelian 2-group with an elementary abelian subgroup $B$ of index 2. If $|B|=2^m$ then what can we say about the order of automorphism group of $A$?

Note that all such $A$ can be constructed via an automorphism of order 2 of $B$, and letting this element square to a fixed point in $B$. I'll try and say more when I get home tonight. — , Aug 23 '13 at 20:52

score 0 · Answer 1 · answered Aug 23 '13 at 20:13

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Not 100% sure on this but here goes: We know any automorphism of $A$ must restrict to an automorphism of $B$, and $Aut(B) = GL(m,2)$. Also, there is a unique nonzero element of $A$ that is in the image of the map $x \rightarrow 2x$. This element must get mapped to itself. So with a suitable basis any automorphism of $A$ restricts to the subgroup of $GL(m,2)$ with the first column $(1,0,0,...0)^t$. This subgroup has size $(2^m-2)(2^m-4)...(2^m-2^{m-1})$. Now in order to extend one of these suitable $B$-automorphisms to an $A$-automorphism it is enough to specify where one of the elements of order 4 goes, there are $2^m$ choices. So I'm getting $|Aut(A)|=(2^m-2)(2^m-4)...(2^m-2^{m-1})2^m$.

answered Aug 23 '13 at 20:13

Nate

11,206

Is it obvious that $B$ is characteristic? – Tobias Kildetoft Aug 23 '13 at 20:14
But in a special case, $A$ can be an elementary abelian group of order $2^{m+1}$... – student Aug 23 '13 at 20:20
Try this reasoning with $A=D_8$ to see what goes wrong. – Aug 23 '13 at 20:48

Order of automorphism group of an almost elementary abelian group

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