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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 \begin{equation}G\xrightarrow{\text{quotient}}G/\operatorname{ker}(\phi)\simeq \operatorname{Im}(\phi)\hookrightarrow G'. \end{equation}

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 \begin{equation}\frac{K}{H\cap K}\simeq \frac{HK}{H}. \end{equation}

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 \begin{equation}\frac{G}{K}\simeq \frac{G/H}{K/H}. \end{equation}

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 Answer 1

up vote 1 down vote accepted

I would interpret it for abstract algebraic structures: in place of 'normal subgroups' we say 'congruence relation' (i.e. equivalence relations preserving all given operations), and in place of 'subgroups' we say 'subalgebras'. In particular, these all hold for sets, subsets and equivalence relations:

  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. $$[a]_\eta \, (\kappa/\eta)\, [a_1]_\eta \iff \exists a',a_1': a\,\eta\, a'\,\kappa\,a_1'\,\eta\, a_1. $$
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Yes. What you said is definitely true. But I am looking for something more concrete (but nontrivial). –  Hui Yu Dec 4 '12 at 15:55

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