# Correspondence Theorem in Group Theory

1. Aug 17, 2013

### TheShrike

Hello,

I'm following the proof for this theorem in my textbook, and there is one part of it that I can't understand. Hopefully you can help me. Here is the part of the theorem and proof up to where I'm stuck:

Let $N$ be a normal subgroup of a group $G$. Then every subgroup of the quotient group $G/N$ is of the form $H/N$ for some subgroup $H$ of $G$ with $H\le G$.

...

Proof: Let $H^*$ be a subgroup of $G/N$ so that it consists of a certain set $\{hN\}$ of left cosets of $N$ in $G$. We define the subset $\beta(H^*)$ of $G$ to be $\{g\in G:gN\in H^*\}$. Then $\beta(H^*)$ clearly contains N and is a subgroup of G:

[here follows demonstration that $\beta(H^*)$ is a subgroup of G]

The portion in red is what I'm having trouble with. I don't see why it's immediately clear that $N$ is contained within $\beta(H^*)$.

Any help is appreciated.

I first thought that perhaps $H^*$ is supposed to correspond to the $H$ of the hypothesis, but then I realised that cannot be true, since the elements of $\beta(H^*)$ would be precisely those of $H$ and the properties would be trivial. In particular, proving that $\beta(H^*)$ is a subgroup of $G$ would be pointless.

2. Aug 18, 2013

### Stephen Tashi

Isn't the identity of $H^*$ exactly the coset $N$?

3. Aug 18, 2013

### TheShrike

That's correct. So we know the identity element $1$ belongs to $\beta(H*)$. We also know that $N$ is a subgroup, so $1\in N$. Therefore $\beta(H*)$ contains at least one element of $N$.

4. Aug 18, 2013

### Stephen Tashi

For each $g \in N ,\ gN = N$ and $N \in H^*$, so $g\in \beta(H^*)$

5. Aug 19, 2013

### TheShrike

Ah, well that clears that up. Thanks.