1. ## Group isomorphism

Hello. Need some fast help on this:

Let $\phi G, \star)\rightarrow(H, \triangle) " alt="\phi G, \star)\rightarrow(H, \triangle) " /> be an isomorphism between these two groups. If $a' \in G$ is the inverse element of $a \in G$, then prove that $\phi(a')\in H$ is the inverse element of $\phi(a) \in H$.

I tried like this:
Since $\phi$ is isomorphism, that means it is a one to one map thus
$\phi(a)=\phi(b)$ only when $a = b$ and since $\phi$ is isomorphism, $\phi(a')$ must be the inverse of $\phi(a)$.

Do you think my prove is enough, if not please elaborate.

2. Originally Posted by javax
Hello. Need some fast help on this:

Let $\phi G, \star)\rightarrow(H, \triangle) " alt="\phi G, \star)\rightarrow(H, \triangle) " /> be an isomorphism between these two groups. If $a' \in G$ is the inverse element of $a \in G$, then prove that $\phi(a')\in H$ is the inverse element of $\phi(a) \in H$.

I tried like this:
Since $\phi$ is isomorphism, that means it is a one to one map thus
$\phi(a)=\phi(b)$ only when $a = b$ and since $\phi$ is isomorphism, $\phi(a')$ must be the inverse of $\phi(a)$.

I can't see any explanation or justification here...in fact, the argument is true for ANY group homomorphism, isomorphism or not, and the proof is boringly simple:

$\phi(a)\phi(a')=\phi(aa')=\phi(e_G)=e_H$

Tonio

Do you think my prove is enough, if not please elaborate.