Can B x A Form a Group with This Operation?

[fk378]

Homework Statement

Let A, B be groups and theta: A --> Aut(B) a homomorphism. For a in A denote theta(a)= theta_a in Aut(B). Equip the product set B x A={(b,a): a in A, b in B} with the binary operation (b,a)(b',a')= (b'',a'') where a''=aa' and b''=b(theta_a{b')).

Show that this binary operation induces a group structure on the set B x A (ie it satisfies the group axioms).

The Attempt at a Solution

How do I show that there exists inverses, an identity element and that it is closed? I tried first for identity:

WTS there exists e such that ae=a=ea. Then I don't know where to go from there. It seems like I am just assuming that there exists e such that aa'=a(a-inverse)=e

Then from there I can show that there exists an inverse right?

 

[Unassuming]

What is aut(B)?

 

[fk378]

Aut(B) is the set of all automorphisms of G. It is a subgroup of A(G), the set of all 1-1 and onto mappings of G onto itself.

 

[Hurkyl]

How do I show that there exists inverses, an identity element

With this sort of problem it's usually easiest to simply write down an explicit formula for them. The formula usually isn't too hard to guess, and when it's not obvious, it can usually be determined by solving the appropriate equation.

and that it is closed?

Again, direct calculation -- this one is usually trivial.

 

[fk378]

That's where I am stuck--writing the formula. Is the one I wrote with my question completely off?

 

[fk378]

Am I supposed to prove there exists an identity or just that the identity is unique?

 

[adriank]

All you need to prove are the group axioms, which are (besides that the binary operation is closed) associativity and the existence of an identity and inverses. From those you can prove that the identity and inverses are unique.

Closure is easy: if (b, a) and (b', a') are in B x A, then show that (b, a)(b', a') is also in B x A by direct calculation. Proving associativity is usually tedious but straightforward; remember that θ and θ_a are homomorphisms.

For the identity and inverses you can either guess, or write an equation. You could guess (correctly) that the identity is (e, e) (with, of course, the first e being the identity of B and the second from A) and prove that it works, or you could suppose, say, that (b_e, a_e) is the identity, write (b, a)(b_e, a_e) = (b, a), calculate the product, and find a_e and b_e. (Don't forget to prove that the multiplication works the other way, i.e. (b_e, a_e)(b, a) = (b, a).) For inverses, in this case the obvious guess at the inverse of (b, a) doesn't work, but you can still find (b, a)^-1 by letting, say, (b, a)^-1 = (b*, a*) and writing (b, a)(b*, a*) = (e, e).