First, I’ll prove the first isomorphism theorem (in case you aren’t already familiar with it).

Statement: If you have a group homomorphism \(\phi : A \to B\), then let \(K = \mathrm{ker}(\phi)\). Then we have that \(A/K \cong \mathrm{im}(\phi) \leq B\).

Proof: Define \(\theta : A/K \to B\) such that \(\theta(gK) = \phi(g)\). This is well-defined, since if \(gK = hK\) (as sets), then \(h^{-1}gK = K\), so \(h^{-1}g \in K\). Therefore, we know that \(\phi(h^{-1}g) = e\), so \(\phi(h) = \phi(g)\), so in particular \(\theta(hK) = \theta(gK)\) when \(gK = hK\).

To show that \(\theta\) is an isomorphism, we write \(\theta(gKhK) = \theta(ghK)\) (since \(K\) is a normal subgroup, \(hK=Kh\)), and \(\theta(ghK) = \phi(gh)\) by definition, which is precisely \(\theta(gK)\theta(hK)\).

Then we need to show \(\theta\) is an isomorphism. We begin by showing it is surjective. If \(h \in \mathrm{im}(\phi)\), then for some \(g \in A\), we have \(\phi(g)=h\). So \(\theta(gK) = \phi(g)=h\). So \(\theta\) is surjective.

Then also we need to show that \(\theta\) is injective. If \(\theta(gK) = \theta(hK)\), then \(\phi(g) = \phi(h)\), so \(h^{-1}g \in K\). So

\[\begin{align*} hK &= h \cdot h^{-1}gK\\ &= gK. \end{align*}\]

So \(\theta(hK)=\theta(gK)\) implies \(hK = gK\), and \(\theta\) is injective. \(\Box\)

Poem (forgive the occasional slant rhyme and be generous with the meter):

We’re given \(\phi\), a map from a group \(A\) to \(B\)
And \(K\) the elements mapping to just the identity.
Then we know that \(K\)’s a normal subgroup of \(A\)
So we obtain a new group when we quotient by \(K\).

Then we make a new function, \(\theta\), with some anticipation
sending cosets of \(K\) to \(\phi\) of some coset representation.
This maps from our new group to the image of \(\phi\),
But how do we know that \(\theta\) is well defined?

If \(gK\) and \(hK\) are two names for the same set
Consider \(h^{-1}\) times \(g\) and what do we get?
An element that lies in \(K\), so that when \(\phi\)’s applied
we get back only the identity, and nothing besides.

Then \(\phi(h^{-1})\) times \(\phi(g)\)
must also yield nothing but \(e\).
So \(\phi(h)\) is \(\phi(g)\), now look how \(\theta\)’s constructed,
its well-definition can no longer be distrusted.

But is it a homom, this function that’s at least defined well?
Naturally, the group operation’s preserved, can’t you tell?
Have \(\theta(gKhK)\) equals \(\theta(ghK)\) by normality
equals \(\phi(gh)\), by our definition’s legality.

Which is \(\phi(g)\) times \(\phi(h)\), since \(\phi\) is given
to be exactly a group homomorphism.
Then this we can write as \(\theta(gK)\) times \(\theta(hK)\)
which makes \(\theta\) a homom, wouldn’t you say?

What’s more, the image of \(\phi\) must be the image of \(\theta\)
I don’t want to be just a definition repeater
so take \(\phi(g)\) equals \(h\), for some \(h\) in im(\(\phi\))
then \(\theta(gK)\) is a clear thing to try.

Now if \(\theta(aK)\) gives us just \(e\)
then \(\phi(a)\) too gives the identity
so \(a\) is in \(K\), then \(aK\)’s just \(K\), and we’re done
since now it’s clear \(\theta\) is one-to-one.

And now we’ve reached the end of my maths lyricism
Since we’ve shown \(\theta\) is an isomorphism.