Let A,B be groups, theta:A-->Aut(B) a homomorphism. Then

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rawkerrxx

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Prove that this is a group

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).



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?
 
Re: Prove that this is a group

This is not an easy, straight-forward problem and takes time to analyze. That said, here is my attempt :) :

Automorphisms are isomorphisms (essentially just a permutation of the elements). Since B is a group, it has an idenity, inverse, etc. Aut(B) is a group under composition.

For identity, you are looking for an element b' and a' that has the following properties:

(1) \(\displaystyle \theta_{a'}(b') \rightarrow e_B\)
(2) \(\displaystyle a' \rightarrow e_A\)

Since \(\displaystyle \theta\) is a homomorphism, the identity \(\displaystyle e_A\) in A gets sent to the identity mapping in Aut(B). Note if we use this fact and (2) as a guide, we have a strong candidate for an identity: \(\displaystyle e = (e_B, e_A)\).

So, \(\displaystyle (b,a)(e_B,e_A) = (b \cdot \theta_{e_A}(e_B), a \cdot e_A) \,\, ...\). Think you could finish? (Don't forget to check right and left inverse).

Once we know the identity, finding inverses gets a little easier.

We need to find a b' and a' s.t.:

(3) \(\displaystyle b \cdot \theta_{a'}(b') =e_B\)
(4) \(\displaystyle a \cdot a' = e_A\)

It seems our hands our tied as to what a' must be from (4) (a-inverse, right?). Now for b'... We now know that we need to find a b' such that:

(5) \(\displaystyle b \cdot \theta_{a^{-1}}(b') = e_B \,\, \,\, \Rightarrow \,\, \,\, \theta_{a^{-1}}(b') = b^{-1}\)

And we have NO CONTROL over that subscript! Good thing we're working with a group of automorphisms on B :wink:

Lets take a gander at the the element in Aut(B) that looks like \(\displaystyle \theta_a\). One property of homomorphisms we certainly like here is the following:

\(\displaystyle f(a^{-1}) = f(a)^{-1}\).

What then happens if we let \(\displaystyle b' = \theta_a(b^{-1})\).

Note that this candidate for b' is certainly in B, as \(\displaystyle \theta_a\) is an Automorphism on B. Think you could peice this together? I have not done the checks myself, so you'd do well to check my work yourself.

Fun problem though :D
 
Re: Prove that this is a group

We're given an element (b,a), to which we need to find an inverse. After that point, a and b are free game to use. I defined b' that way for a reason.

Just see that \(\displaystyle \theta_a \circ \theta_{a^{-1}} = id_B\), where \(\displaystyle id_B\) is the idenity mapping/automorphism on B.

edit: To make my last post flow more smoothly at that point (which I honestly haven't realized until now), I could have said, "apply the automorphism \(\displaystyle \theta_a\) to both sides of the RHS of [5]."
 
Re: Prove that this is a group

Ah, I see. Okay, I think I understand now what you've proved. Thank you! I've proved closure myself as well. Now I don't know how to show associativity. Can you give any hints?
 
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')).

(a) Assume that p,q in N are prime and p divides (q-1). Consider the case A=Zmodp, B=Zmodq. Show that there exists phi in Aut(Zmodq) which has order p.

Hint: Use Cauchy's Thm for Abelian groups

(b) Deduce that for any 2 primes p,q in N such that p|(q-1) there is a non-Abelian group of order pq.

------

(a) If p divides q-1, then p=-1 in mod q. Don't really know where to go from there.....

(b) No idea.
 
Re: Prove that this is a group

Have you attempted it yourself?

[(b1,a1)*(b2,a2)] * (b3,a3) = ??

(b1,a1) * [(b2,a2)*(b3,a3)] = ??
 
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