# Thread: Show that a group is isomorphic to another

1. ## Show that a group is isomorphic to another

Let A and B be groups. Show that $N = \{ (a,1) \, | \, a \in A \}$ is a normal subgroup of $A \times B$ and the quotient of $A \times B$ by this subgroup is isomorphic to B
I've shown that it is a normal subgroup, but I don't know about the isomorphic part. I know you have to show that it is a homomorphism, surjective and injective, but I don't really know how to do that in this case.

2. ## Re: Show that a group is isomorphic to another

Hi,
Do you know the "fundamental" theorem of homomorphisms for groups? That is given $\phi:G\rightarrow H$ a homomorphism from the group G onto the group H, then ${G\over Ker(\phi)}\cong H$ where $Ker(\phi)=\{g\in G:\phi(g)=1\}$.

If so, define $\phi:A\times B\rightarrow B$ by $\phi(a,b)=b$ for all $(a,b)\in A\times B$. Then apply the fundamental theorem.

If you don't know this theorem, define the "map" $\psi:{A\times B\over N}\rightarrow B$ by $\psi(a,b)N=b$. Show that $\psi$ is well defined (if $(a_1,b_1)N=(a_2,b_2)N$, then $b_1=b_2$) and that in fact $\psi$ is an isomorphism.

3. ## Re: Show that a group is isomorphic to another

Recall that $(a,b)N = (a',b')N$ if and only if $(a,b)\ast[(a',b')]^{-1} \in N$.

Now:

$(a,b)\ast[(a',b')]^{-1} = (a,b)\ast(a'^{-1},b'^{-1}) = (aa'^{-1},bb'^{-1})$.

Since $A$ is a group, we certainly have $aa'^{-1} \in A$, so $(aa'^{-1},bb'^{-1}) \in N = A \times \{1_B\}$ if and only if $bb'^{-1} = 1_B$, and thus $b = b'$.

It should be clear that $(a,b)N \mapsto b$ is a homomorphism (it's a straightforward computation).

Also $(1_A,b)N$ is clearly a pre-image for $b$.

So all you need to do is show $\text{ker }\psi = N$ (hint: $(a,1_B)N = N$).

4. ## Re: Show that a group is isomorphic to another

Originally Posted by johng
Hi,
Do you know the "fundamental" theorem of homomorphisms for groups? That is given $\phi:G\rightarrow H$ a homomorphism from the group G onto the group H, then ${G\over Ker(\phi)}\cong H$ where $Ker(\phi)=\{g\in G:\phi(g)=1\}$.

If so, define $\phi:A\times B\rightarrow B$ by $\phi(a,b)=b$ for all $(a,b)\in A\times B$. Then apply the fundamental theorem.

If you don't know this theorem, define the "map" $\psi:{A\times B\over N}\rightarrow B$ by $\psi(a,b)N=b$. Show that $\psi$ is well defined (if $(a_1,b_1)N=(a_2,b_2)N$, then $b_1=b_2$) and that in fact $\psi$ is an isomorphism.
Why do I have to show that it is well defined? I thought the question asks for isomorphism (which is injective, surjective and homomorphism)

Originally Posted by Deveno
Recall that $(a,b)N = (a',b')N$ if and only if $(a,b)\ast[(a',b')]^{-1} \in N$.

Now:

$(a,b)\ast[(a',b')]^{-1} = (a,b)\ast(a'^{-1},b'^{-1}) = (aa'^{-1},bb'^{-1})$.

Since $A$ is a group, we certainly have $aa'^{-1} \in A$, so $(aa'^{-1},bb'^{-1}) \in N = A \times \{1_B\}$ if and only if $bb'^{-1} = 1_B$, and thus $b = b'$.
Again, why are you proving the function is well defined?

It should be clear that $(a,b)N \mapsto b$ is a homomorphism (it's a straightforward computation).
Homomorphism, got it

Also $(1_A,b)N$ is clearly a pre-image for $b$.
Surjective, got it

So all you need to do is show $\text{ker }\psi = N$ (hint: $(a,1_B)N = N$).
Sorry, what does this do? Can you be more explicit, like say what you are trying to show at each step?

5. ## Re: Show that a group is isomorphic to another

When we have a map from a quotient group to another group defined in terms of coset representatives, we need to be sure the map is CONSTANT on all elements of the coset.

In other words it's not simply enough to define $\psi((a,b)N) = b$, because $\psi$ should depend ONLY ON THE COSET, and not $a$ and $b$.

To make sure this is true, we need to check if for any other pair $(a',b')$ such that $(a,b)N = (a',b')N$ we get the same value for $\psi$.

To answer your last question, we have:

Theorem: A group homomorphism $\phi:G \to G'$ is injective if and only if $\text{ker }\phi = \{e_G\}$.

(Recall that $\text{ker }\phi = \{g \in G: \phi(g) = e_{G'}\}$).

Suppose that $\phi$ is injective. This means that if $\phi(g) = \phi(h)$ then $g = h$. In particular, for any $g \in \text{ker }\phi$ we have $\phi(g^2) = \phi(g)\phi(g) = e_{G'}e_{G'} = e_{G'} = \phi(g)$,

so we conclude from injectivity that for any $g$ in the kernel, $g^2 = g$. Multiplying both sides of this last equation (on either side) by $g^{-1}$ gives $g = e_G$, that is, any element of the kernel is the identity of $G$, that is: $\text{ker }\phi = \{e_G\}$.

Now suppose that the kernel is just the identity of $G$. If we have $g,h \in G$ such that $\phi(g) = \phi(h)$ then $\phi(g)(\phi(h))^{-1} = e_{G'}$.

SInce $\phi$ is a homomorphism, $e_{G'} = \phi(g)(\phi(h))^{-1} = \phi(g)\phi(h^{-1}) = \phi(gh^{-1})$, so $gh^{-1} \in \text{ker }\phi$.

But the kernel is just the identity of $G$, so $gh^{-1} = e_G$. Multiplying both sides of THIS equation on the right by $h$ gives $g = h$, so $\phi$ is injective, which concludes the proof.