# Intuition about the first isomorphism theorem

I'm currently studying group theory and recently I've read about the first isomorphism theorem which can be stated as follows:

Let $G$ and $H$ be groups and $\varphi :G\to H$ a homomorphism, then $\ker \varphi$ is a normal subgroup of $G$, $\varphi(G)$ is a subgroup of $H$ and $G/\ker \varphi \simeq \varphi(G)$.

The proof is quite easy, but I've been thinking about what's the best way to understand this result. In that setting I've came up with the following intuition:

It's easy to see that a homomorphism $\varphi : G\to H$ is injective if and only if $\ker \varphi = \{e\}$ where $e$ is the identity of $G$.

Now, my intuition about the first isomorphism theorem is: if $\varphi : G\to H$ is a homomorphism which is not injective, we can then construct a new group on which the equivalent homomorphism is indeed injective. We do this by quotienting out what is in the way of making $\varphi$ injective, that is, everything that is in the kernel.

In that way taking the quotient $G/\ker \varphi$ we construct a group on which we "kill" everything that is in the kernel of $\varphi$. The natual projection of $\varphi$ to this quotient will then be an injective function.

So is this the best way to understand the first isomorphism theorem? It's a way to "get out of the way" everything which is stoping a homomorphism from being an injective map? If not, what is the correct intuition about this theorem and its importance?

• I actually always liked the intuition of thinking as every element as a translation of your identity. More concretely, think of your kernel, and "divide" it out of your group G. This effectively forces you create a new group with the same cardinality as the image of $\phi$, and so of course you can now create a bijection since you're already given a homomorphism. I highly recommend Dummit and Foote's discussion of this in the language of "fibers". – Rellek Mar 13 '16 at 22:31
• I think you got the idea perfectly. I began reading the question, and I knew exactly what I was going to answer... then I continue reading the question, and you say everything yourself. Come on :P – Ivo Terek Mar 13 '16 at 23:00

I understand the Theorem in the same way as you. The idea with a lot of these Algebra Theorems, where we factor an algebraic structure through a quotient, is to remove some undesirable part of the structure.

In this case, we want to get an isomorphism out of a surjective homomorphism, which is a much "nicer" map. So, we quotient out by the kernel, and as a result we have a map where only the zero element is sent to zero, which maintains surjectivity.

This sort of Theorem reappears frequently, and yours is the correct intuition.

The fiber point of view is the one I like, because it captures the idea that when you quotient out $ker\phi$, you identify through $\sim$ all the points that $\phi$ sends to $0$.

To extend this a bit, suppose we take a topological space $X$, a space $Y$ and an $f:X\to Y$, and topologize $Y$ by declaring that $V$ open in $Y$ $\Leftrightarrow f^{-1}(V)$ open in $X.$

Now given any $\sim$ on $X$, $q:X\to X/\sim$ induces a topology on $X/\sim$ as above and from this you get the following result:

if $g:X\to Z$ is continuous, then there is a unique $\ \overline g:X/\sim \to Z$ such that $\overline g\circ q=g$

and so we have a topological analog of the First Isomorphism Theorem.