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Problem 103 (Subgroups Of A Quotient Group Correspond To Subgroups Of The Original Group)

Suppose that $N$ is a normal subgroup of $G$ and that $B$ is a subgroup of $G/N$. Prove the following:

  1. There exists a subgroup $H\leq G$ such that $H/N=B$.
  2. If we know that $n=|N|$ and $m=|B|$ (so $n$ is the order of $N$ in $G$, and $m$ is the order of $B$ in $G/N$), then $G$ must have a subgroup $H$ of order $nm$.

As a suggestion, consider the homomorphism $f:G\to G/N$ given by $f(g)=Ng$ and use some properties of homomorphisms.

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