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January 8[edit]

Group Theory Question[edit]

Does anyone know the status of the following question: Given a finite group $G$ and a subgroup $H$ of $G$ , is it true that the automorphism group of $H$ , $Aut(H)$ isomorphic to a subgroup of the automorphism group of $G$ , $Aut(G)$ ? My initial instinct was that this is false, but I haven't been able to produce a counterexample yet (admittedly, I haven't really put a lot of effort into that search yet). 76.14.232.104 (talk) 12:47, 8 January 2015 (UTC)[reply]

In order for it to be a subgroup it must, to begin with, be a subset. It isn't because elements of

Aut H

are maps from

H

to

H

while elements of

Aut G

aren't. YohanN7 (talk) 13:10, 8 January 2015 (UTC)[reply]

The real question is if you can extend any element of

Aut H

in such a way that it is an element of

Aut G

and such that group multiplication in

Aut G

is compatible with the extension. YohanN7 (talk) 13:25, 8 January 2015 (UTC)[reply]

... or one can read the question as "is isomorphic to a subgroup of" instead of "is a subgroup of", which is the way I think it would normally be read when considering questions of this nature (one is rarely interested in more than isomorphism here). I guess the OP's initial reaction is that because the structure of the group H might have automorphisms that do not extend to the full group G. While I cannot answer the question, if hunting for counterexamples, a comment in Automorphism#Examples hints that it might make sense to examine a subgroup H with a nontrivial centre. —Quondum 14:45, 8 January 2015 (UTC)[reply]

The isomorphism formulation may lose too much information. Consider a small subgroup of some infinite group, like a two-element group. Ah, finite groups. Well, the objection remains. A subgroup may occur many times in a containing group. YohanN7 (talk) 15:30, 8 January 2015 (UTC)[reply]

Consider e.g., G the symmetric group on 6 symbols, H the subgroup generated by the 2-cycles (12), (34)(56). There is an automorphism of H interchanging the two generators, but this is not induced by any automorphism of G. Sławomir Biały (talk) 16:11, 8 January 2015 (UTC)[reply]

My original intent was that it only be isomorphic to a subgroup of $Aut(G)$ , I've edited the OP to reflect that. 76.14.232.104 (talk) 01:35, 9 January 2015 (UTC)[reply]

Let G be C₄×C₂ and H the subgroup isomorphic to C₂×C₂. Then #Aut(G) = 8 and #Aut(H) = 6 and the orders preclude Aut(G) from having an isomorphic copy of Aut(H). --RDBury (talk) 07:43, 9 January 2015 (UTC)[reply]

Number of edges in an nxnxn cube[edit]

The formula for the number of edges in an nxn square is 2n(n+1). What's the formula for the number of edges in an nxnxn cube?? I know that for n=1, the answer is 12. For n=2, I know that:

There are 27 vertexes, but I want to know the number of edges.
The number is bounded below by 36, because there are 36 edges on a group of three 2x2 squares.

Georgia guy (talk) 17:42, 8 January 2015 (UTC)[reply]

I'm pretty sure it's

3n(n+1)^{2}.

There are 3 ways to choose the direction,

(n+1)^{2}

ways to choose the location of the ray, and

n

ways to choose the segment along the ray.

This generalizes to

dn(n+1)^{d-1}

for a d-dimensional hypercube. -- Meni Rosenfeld (talk) 17:57, 8 January 2015 (UTC)[reply]