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I'm trying to prove the correspondence theorem differently than my professor did. Is this valid?

Theorem: Let $G$ be a group and $N \le G$ be a normal subgroup. Then there is a $1-1$ correspondence between subgroups of $G/N$ and subgroups of $G$ containing $N$.

Proof: Let *Sub*$ (G/N)$ and *Sub*$(G)$ denote the set of subgroups of $G/N$ and subgroups of $G$ containing $N$ respectively. Define the map $f$: *Sub*$(G)$ $\to$ *Sub*$(G/N)$ by $f(H) = H/N$. This map is injective since if $f(H) = f(H')$, by the fact that the cosets partition a group,

$$H = \bigcup_{a \in H} aN = \bigcup_{a \in H'} aN = H'$$

To show that $f$ is surjetive, let $K \le G/N$ be a subgroup. We show that $V = \cup_{\alpha \in K} \alpha$ is a subgroup. Let $x,y \in V$. It follows that $x = an $ and and $y = bn'$ where $a,b \in G$ and $n, n' \in N$. So $xy = (an)(bn') = abn''n$ for some $n'' \in N$ , and hence $xy \in V$. Furthermore, for any $an \in V$ $(an)^{-1} =n^{-1}a^{-1} \in V$, so $V$ is a subgroup which clearly contains $N$, and $f(V) = K$.

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1 Answer 1

You get into trouble here:

To show that $f$ is surjetive, let $K \le G/N$ be a subgroup. We show that $V = \cup_{\alpha \in K} \alpha$ is a subgroup. Let $x,y \in V$. It follows that $x = an $ and and $y = bn'$ where $a,b \in G$ and $n, n' \in N$. So $xy = (an)(bn') = abn''n$ for some $n'' \in N$ , and hence $xy \in V$. Furthermore, for any $an \in V$ $(an)^{-1} =n^{-1}a^{-1} \in V$, so $V$ is a subgroup which clearly contains $N$, and $f(V) = K$.

The fact that you never used the hypothesis that $K$ is a group should make you very suspicious. A minor point first: normality of $N$ in $G$ gives you $(an)(bn')=abn''n'$ for some $n''\in N$, not $abn''n$. (You’re using $nb=bn''$.) Let’s assume that that’s been fixed, so that you know that $xy=abm$ for some $m\in N$ and $a,b\in G$; how does that tell you that $xy\in V$?

In order to show that $xy\in V$, you must find an $\gamma\in K$ such that $xy\in\gamma$. This can be done, but you have to work a bit more. Start with the assumption that $x,y\in V$: this means that there are $\alpha,\beta\in K$ such that $x\in\alpha$ and $y\in\beta$, and there are $a,b\in G$ such that $\alpha=aN$ and $\beta=bN$. Now pick your $n_x,n_y\in N$ such that $x=an_x$ and $y=bn_y$, so that $xy=an_xbn_y=abnn_y$ for some $n\in N$ by normality of $N$ in $G$. Clearly, then, $xy\in abN$, and we now want to show that $abN\in K$. The obvious candidate is $\alpha\beta$, so the natural things to do next is to try to prove that $abN=\alpha\beta$. But that’s just proving that $abN=(aN)(bN)$, which you presumably already proved when constructing quotient groups in the first place, so we’ve shown that $V$ is closed under the group operation.

Your argument for closure under taking inverses has the same problem. You can’t say that $n^{-1}a^{-1}\in V$ unless you can show that there is some $\gamma\in K$ such that $n^{-1}a^{-1}\in\gamma$. If $a\in\alpha\in K$, the obvious candidate is $\alpha^{-1}$, where the inverse is taken in $G/N$, and you can finish off the argument much as I did the one for closure under the operation.

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Would this prove closure under inverse? Let $an \in V$. Since $aN \in K$, $a^{-1}N = Na^{-1} \in K$ and so $n^{-1}a^{-1} \ in V$ –  Jonah777 Sep 23 '12 at 6:44
    
@1111: It would be better to be a little more explicit: If $a\in V$, then $a\in\alpha$ for some $\alpha\in K$. Since $a\in aN$, we can even say that $aN=\alpha\in K$, so $a^{-1}N=(aN)^{-1}=\alpha^{-1}\in K$, since $K$ is a subgroup of $G/N$. And $a^{-1}\in a^{-1}N$, so $a^{-1}\in V$. –  Brian M. Scott Sep 23 '12 at 6:46

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