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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?

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    $\begingroup$ 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". $\endgroup$ – Rellek Mar 13 '16 at 22:31
  • $\begingroup$ 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 $\endgroup$ – Ivo Terek Mar 13 '16 at 23:00
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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.

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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.

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