# Understanding the three isomorphism theorems

I have learnt the following three isomorphisms for a while but without true understanding:

A group homomorphism $\phi:G\to G'$ can be decomposed into $$G\xrightarrow{\text{quotient}}G/\operatorname{ker}(\phi)\simeq \operatorname{Im}(\phi)\hookrightarrow G'.$$

and

$H$ is a normal subgroup of $G$ and $K$ is another subgroup. Then $H\cap K$ is normal in $K$, $HK$ is a subgroup inside which $H$ is normal, and $$\frac{K}{H\cap K}\simeq \frac{HK}{H}.$$

and

$H$ is a subgroup $G$ and $K\supset H$ is another subgroup. Then $K/H$ is normal in $G/H$ if and only if $K$ is normal in $G$. If $K$ is normal then $$\frac{G}{K}\simeq \frac{G/H}{K/H}.$$

The proofs for these three theorems are rather straightforward, and after teaching myself some category theory I am more comfortable with the first one. But I do not feel them. (Like in this post by Gowers he explains Orbit-Stablizer by moving a cube and with this picture you get the feeling that such a theorem has to be right.)

I wonder whether someone can share similar insights on the three isomorphisms maybe by using intuitive-but-nontrivial examples like Gowers.

Thanks!

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A pedantic point: You may wish to fix your last quotation. It is quoting what you, I think, mean to represent your own words. –  000 Dec 2 '12 at 16:43
@Limitless Thanks! Done. –  Hui Yu Dec 2 '12 at 16:48
The second one is best seen with a diagram: opensourcemath.org/books/gaglione-gp-thry/img1818.png I cannot give you an intuitive description of the third. However, I found that trying to prove it myself helped me understand it. –  user1729 Dec 3 '12 at 15:53
@user1729 I do not quite understand the diagram. What does it say? –  Hui Yu Dec 4 '12 at 15:54
@HuiYu: It is a subdiagram of the subgroup lattice diagram. From it you can see that $H_2<H_1H_2$ and $H_1\cap H_2< H_1$, and you get $H_1H_2/H_2\cong H_1/(H_1\cap H_2)$ because they correspond to the same "line" in the diagram, shifted diagonally downwards. The other wise of a square, if you will. –  user1729 Dec 12 '12 at 9:58

1. Any homomorphism $\phi:A\to B$ can be decomposed into $A\twoheadrightarrow A/\ker \phi \cong im\phi\hookrightarrow B$, where $im\phi$ is the range and $$a\,(\ker\phi)\,a_1 \iff f(a)=f(a_1).$$
2. Let $\eta$ (in place of $H$) be a congruence on an algebra $A$, and $B$ (in place of $K$) be a subalgebra of $A$, then $H\cap K$ corresponds to the congruence $\eta|_B\,(:=\eta\,\cap\, B\times B)$ on $B$, and $HK$ corresponds to the subalgebra $\eta(B):=\{a\in A\mid \exists b\in B:\, a\,\eta\, b\}$.
3. Now $K$ and $H$ both wanted to be normal subgroups, so in place of them we have congruences $\kappa$ and $\eta$ on an algebra $A$, and we consider $\kappa/\eta$ which is the induced congruence on $A/\eta$ by $\kappa$, i.e. $$[a]_\eta \, (\kappa/\eta)\, [a_1]_\eta \iff \exists a',a_1': a\,\eta\, a'\,\kappa\,a_1'\,\eta\, a_1.$$