# Subgroup of factor groups

Printable View

• Oct 20th 2007, 08:52 PM
tttcomrader
Subgroup of factor groups
Let G be a group, and let H be normal in G. Prove that every subgroups of G/H is of the form K/H, where K is a subgroup of G with $\displaystyle H \subset K$.

My proof so far:

Let $\displaystyle Y:G \rightarrow \frac{G}{H}$ , let X be a subgroup of G/H. By a theorem, $\displaystyle Y^{-1}(X) = \{t \in \frac{G}{H} : Y(t) \in X \}$ is a subgroup of G.

Any help from here?

Thanks.
• Oct 21st 2007, 07:38 AM
ThePerfectHacker
Quote:

Originally Posted by tttcomrader
Let G be a group, and let H be normal in G. Prove that every subgroups of G/H is of the form K/H, where K is a subgroup of G with $\displaystyle H \subset K$.

My proof so far:

Let $\displaystyle Y:G \rightarrow \frac{G}{H}$ , let X be a subgroup of G/H. By a theorem, $\displaystyle Y^{-1}(X) = \{t \in \frac{G}{H} : Y(t) \in X \}$ is a subgroup of G.

Any help from here?

Thanks.

I have no idea what it means "of the form K/H"
Given the canonical homormorphism $\displaystyle \gamma: G\mapsto G/H$.
If $\displaystyle L\leq G/H$ is a subgroup, i.e. $\displaystyle L = qH$ then,
$\displaystyle \gamma^{-1} [L] = \{ x\in G | \gamma (x) \in qH \} = \{ x\in G | xH \in qH \} = qH$.