some abstract algebra help needed

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Nov 25th 2006, 06:04 PM
harbong

some abstract algebra help needed

Hi, I have a few questions on my abstract algebra homework that is bugging me, and wondered if anyone can send me some help!

1. Let G, G' and G'' be groups. let φ: G > G' and γ: G' > G' be homomorphisms. Prove γ(φ(x)) is a homomorphism from G to G''.

2. If G is an abelian group, prove that the mapping θ: G > G' defined by θ(x)=x^-1 is an automorphism.

3. Suppose φ:G > G' is an isomorphism from G to G'. Prove that if H is a subgroup of G, then φ(H) (the image set of H under the map) is a subgroup of G''.

any help would be appreciated, proofs always have me stumped!
Nov 25th 2006, 07:12 PM
ThePerfectHacker

Quote:

Originally Posted by harbong

1. Let G, G' and G'' be groups. let φ: G > G' and γ: G' > G' be homomorphisms. Prove γ(φ(x)) is a homomorphism from G to G''.

First the composition $\gamma \circ \phi$ maps $G\to G''$ .
We need to show,
$\gamma \circ \phi (xy)=\gamma \circ \phi(x) \gamma \circ \phi(y)$
Now,
$\gamma \circ \phi(xy)=\gamma (\phi (xy))=\gamma(\phi(x) \phi(y))$ because $\phi$ is a homomorphism, and $\phi(x),\phi(y)\in G'$ thus, $\gamma(\phi(x)) \gamma(\phi(y))$ because $\gamma$ is also a homomorphism. Q.E.D.

Quote:

2. If G is an abelian group, prove that the mapping θ: G > G' defined by θ(x)=x^-1 is an automorphism.

There is a problem with this question.
Because an automorphism is defined between the same groups but you did not mention that. Thus, I assume the groups are the same.
We need to show that,
$\theta: G\to G$ is an isomorpism.
Simple, first this is a function between these two sets.
Next,
$\theta (x)=\theta (y)$
Thus,
$x^{-1}=y^{-1}$
Thus,
$x=y$
This show the map is one-to-one.

Next, for any $g\in G$ we can choose $g^{-1}\in G$ (for it has an inverse) then, $\theta (g^{-1})=g$ . Thus this map is onto.

Finally we show the homomorphism part.
$\theta(xy)=(xy)^{-1}$
In general the inverse for any group is,
$y^{-1}x^{-1}$
But this group is abelian thus,
$x^{-1}y^{-1}=\theta(x)\theta(y)$
Nov 25th 2006, 07:25 PM
ThePerfectHacker

Quote:

Originally Posted by harbong

3. Suppose φ:G > G' is an isomorphism from G to G'. Prove that if H is a subgroup of G, then φ(H) (the image set of H under the map) is a subgroup of G''.

It is not necessary to state the map is a group isomorphism, saying that it is a group homomorphism is sufficient to show that condition.

If,
$H\leq G$
Then,
$\phi [H]=\{\phi (h)|h\in H\}$

Associative: That part is simple, it uses the same binary operation as does $G'$ which is a group and hence is associative.

Identity: We know that, $ge=eg=g$ for all $g\in G$ then its map,
$\phi(ge)=\phi(eg)=\phi(g)$
But because this is a homomorphism we have,
$\phi(g)\phi(e)=\phi(e)\phi(g)=\phi(g)$
Thus we see that $\phi(e)$ is the identity element.

Inverse: Let $\phi(h)\in \phi[H]$ then,
$hh^{-1}=h^{-1}h=e$
Thus, the map,
$\phi(hh^{-1})=\phi(h^{-1}h)=\phi(e)$
But because we have a homomorphism,
$\phi(h)\phi(h^{-1})=\phi(h^{-1})\phi(h)=\phi(e)$
Where, $\phi(e)$ is the identity element.
Thus, we see by definition of an inverse that,
$[\phi(h)]^{-1}=\phi(h^{-1})$ .

Closure: Let $\phi(h_1),\phi(h_2)\in \phi[H]$ .
Then,
$\phi(h_1)\phi(h_2)=\phi(h_1h_2)$
Where, $h_1h_2\in H$ for it is closed.
The by definition of a functional image we see that,
$\phi(h_1h_2)=\phi(h_1)\phi(h_2)\in \phi[H]$
And hence we have,
$\phi[H]\leq G'$
Nov 26th 2006, 09:11 PM
harbong

thank you so much!