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Theorem: Let $G$ and $G'$ be groups and let $f:G\to G'$ be a group homomorphism. Then $G/\textrm{ker}\, f \cong\textrm{im}\, f$.

My question is how to understand this theorem intuitively.

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The most intuitive way I ever got to is "all short exact sequences really look the same". –  Alexei Averchenko Oct 27 '12 at 8:05
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groupprops.subwiki.org/wiki/… –  JavaMan Oct 27 '12 at 8:07
    
But I am thinking what does it mean that we are moding out kernel from G and then it is isomorphic to image how to think about that? –  Reader Oct 27 '12 at 8:10

2 Answers 2

up vote 6 down vote accepted

The intuition is that the group looks similar to its image under the homomorphism when it is divided by a certain subgroup, since distinct elements may get mapped to the same element. By taking the quotient, elements which are mapped to the same are grouped together.

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By why are we just moding out kernel, is it enough to mod out kernel. Because you can also have that $x\mapsto p$ and $y\mapsto p$ what about this, then we are not moding out these elements? Why? –  Reader Oct 27 '12 at 8:18
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@Reader: In that case $xy^{-1} \in \ker f$. –  wj32 Oct 27 '12 at 8:24
    
And since $Ker f$ is a group so we are also moding out xy^{-1} this element. It means that we are moding out all elements which have same image in $G'$. Therefore it make sense that $G/ker\, f\cong im \, f.$ Thanks –  Reader Oct 27 '12 at 8:28
    
Now, I don not understand one thing, why $xy^{-1}\in ker f$ –  Reader Oct 27 '12 at 8:32
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@Reader: $f(xy^{-1})=f(x)f(y)^{-1}=pp^{-1}=e$. Another way of looking at it is to notice that $x$ and $y$ are in the same coset in $G/\ker f$, so they are grouped together because as you said, they have the same image. –  wj32 Oct 27 '12 at 8:40

Maybe it gets more intuitive if you look at the situation with groups replaced by sets. If $f : M \rightarrow N$ is a map between sets $M$ and $N$, then you get an equivalenve relation $R_f := \{(x,y) \in M \times M \::\: f(x) = f(y)\}$. The equivalence class of $x \in M$ is $f^{-1}(\{f(x)\})$, so we have a well-defined (and injective) map from $M/R_f$ (the set of equivalence classe) to $N$, sending $f^{-1}(\{f(x)\})$ to $f(x)$. Thus, first collecting all elements, which map to the same element via $f$, and then mapping these collections to their respective values gives you a bijection onto the image.

Note that in the case of groups $G$ and $G'$ and group homomorphism $f: G \rightarrow G'$ you have $f^{-1}(\{f(x)\}) = x\ker(f)$ for all $x \in G$ and the induced map is a group homomorphism, yielding an isomorphism to the image.

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Nice solution. First we collect elements which have same image and then mapping. –  Reader Oct 27 '12 at 8:37
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Yes. In nice structures just as groups, vector fields, or modules over a ring, the preimage of the unit element (or $0$ if grp operation is $+$) uniquely determines the preimage of all other elements. –  Berci Oct 27 '12 at 10:53
    
Basically, all you need is a Mal'cev operation. –  Zhen Lin Oct 27 '12 at 11:30

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