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Proof of first isomorphism theorem of group

Let $G_1, G_2$ be groups. If $f: G_1 \rightarrow G_2$ is a group homomorphism with $K = \ker(f)$, then $G_1 / K$ is isomorphic to $f(G_1)$.

This was a theorem in the book that was left unproven and I'm really curious as to how you would go about it.

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marked as duplicate by William, DonAntonio, yunone, rschwieb, Gerry Myerson Sep 6 '12 at 12:45

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This is just routine proof, if you know the definition of isomorphism then you can do it –  pritam Sep 6 '12 at 7:55
    
You can find a clear proof in any group theory book of this theorem. :) –  B. S. Sep 6 '12 at 7:56
    
I can't think of a mapping that would make sense intuitively. And this is my first algebra course so nothing is 'routine' for me yet –  Student Sep 6 '12 at 7:56
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There is only one reasonable map to try: $\varphi(aK)=f(a)$. You just have to prove that $\varphi$ is well-defined, meaning that if $aK=bK$, then $f(a)=f(b)$, and that it satisfies the definition of an isomorphism. –  Brian M. Scott Sep 6 '12 at 7:59
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@pritam: Not quite, but almost. –  Brian M. Scott Sep 6 '12 at 8:00
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1 Answer

up vote 1 down vote accepted

You can get this theorem from any starting Algebra book, you can refer Herstein. But why not try yourself?

First prove that $\ker (f)$ is subgroup in $G_1$ (in fact it's a normal subgroup). Then

Define $h: G_1/\ker(f) = G_2$ by $h(g_1 + (\ker f)) = f(g_1)$, and check if this is isomorphism or not.

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Thanks rschwieb for editing, I will start learning Latex. –  Ram Sep 6 '12 at 13:01
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