# Thread: Prove a group is isomorphic to another

1. ## Prove a group is isomorphic to another

Let $\displaystyle G$ and $\displaystyle H$ be groups and let [tex] G^\star [tex] be the subset of $\displaystyle G X H$ consisting of all $\displaystyle (a, e)$ with $\displaystyle a \in G$

Show the following:

1. $\displaystyle G^\star \cong G$
2. $\displaystyle G^\star$ is a normal subgroup of $\displaystyle G X H$
3. $\displaystyle \frac{(G X H)}{G^\star} \cong H$

I showed the first two just fine, but I'm having slight trouble with number 3. I had shown that it's a homomorphism, but the surjective part I'm stuck on. Could anyone guide me through this?

2. For the third one, apply the first isomorphism theorem to the the projection homomorphism $\displaystyle \pi : G\times H \rightarrow H : (g,h) \mapsto h$.

The map $\displaystyle \pi$ should be obviously surjective. For instance $\displaystyle \pi (1, h) = h$ for any given $\displaystyle h$.

3. Originally Posted by crushingyen
Let $\displaystyle G$ and $\displaystyle H$ be groups and let [tex] G^\star [tex] be the subset of $\displaystyle G X H$ consisting of all $\displaystyle (a, e)$ with $\displaystyle a \in G$

Show the following:

1. $\displaystyle G^\star \cong G$
How about $\displaystyle \phi:G\mapsto G^{\star}$ by $\displaystyle \phi(g)=(g,e)$? This is clearly bijective, to see that it's a homomorphism merely note that $\displaystyle \phi(gg')=(gg',e)=(gg',ee)=(g,e)(g',e)=\phi(g)\phi (g')$

2. $\displaystyle G^\star$ is a normal subgroup of $\displaystyle G X H$
We see that given any $\displaystyle (g,e)\in G^{\star}$ and any $\displaystyle (g',h)\in G\times H$ that $\displaystyle \left(g',h\right)\left(g,e\right)\left(g',h\right) ^{-1}=\left(g'g,h\right)\left(g'^{-1},h^{-1}\right)=\left(g'gg'^{-1},hh^{-1}\right)=\left(g'gg'^{-1},e\right)\in G^{\star}$ it clearly follows that $\displaystyle G\lhd G\times H$

4. Originally Posted by Bruno J.
For the third one, apply the first isomorphism theorem to the the projection homomorphism $\displaystyle \pi : G\times H \rightarrow H : (g,h) \mapsto h$.

The map $\displaystyle \pi$ should be obviously surjective. For instance $\displaystyle \pi (1, h) = h$ for any given $\displaystyle h$.
The problem is that I'm not allowed to use the first isomorphism theorem to answer this question. Otherwise, this part would've been trivial.

5. Originally Posted by crushingyen
The problem is that I'm not allowed to use the first isomorphism theorem to answer this question. Otherwise, this part would've been trivial.
Well you should have been specific.

Consider the map $\displaystyle f : (G \times H)/G^* \rightarrow H$ which sends the coset $\displaystyle (h,g)G^*$ to $\displaystyle h$. Show that this map is an isomorphism.