# Thread: some abstract algebra help needed

1. ## 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!

2. 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 $\displaystyle \gamma \circ \phi$ maps $\displaystyle G\to G''$.
We need to show,
$\displaystyle \gamma \circ \phi (xy)=\gamma \circ \phi(x) \gamma \circ \phi(y)$
Now,
$\displaystyle \gamma \circ \phi(xy)=\gamma (\phi (xy))=\gamma(\phi(x) \phi(y))$ because $\displaystyle \phi$ is a homomorphism, and $\displaystyle \phi(x),\phi(y)\in G'$ thus, $\displaystyle \gamma(\phi(x)) \gamma(\phi(y))$ because $\displaystyle \gamma$ is also a homomorphism. Q.E.D.
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,
$\displaystyle \theta: G\to G$ is an isomorpism.
Simple, first this is a function between these two sets.
Next,
$\displaystyle \theta (x)=\theta (y)$
Thus,
$\displaystyle x^{-1}=y^{-1}$
Thus,
$\displaystyle x=y$
This show the map is one-to-one.

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

Finally we show the homomorphism part.
$\displaystyle \theta(xy)=(xy)^{-1}$
In general the inverse for any group is,
$\displaystyle y^{-1}x^{-1}$
But this group is abelian thus,
$\displaystyle x^{-1}y^{-1}=\theta(x)\theta(y)$

3. 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,
$\displaystyle H\leq G$
Then,
$\displaystyle \phi [H]=\{\phi (h)|h\in H\}$

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

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

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

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

4. thank you so much!