Given a normal subgroup \(H\) of a finite group \(G\), is \(G/H\) always isomorphic to a subgroup \(K \le G\)?

### The case of an abelian group

According to the **fundamental theorem of finite abelian groups**, every finite abelian group \(G\) can be expressed as the direct sum of cyclic subgroups of prime-power order: \[G \simeq \bigoplus_{i=1}^u \mathbb{Z}_{p_i^{\alpha_i}}\] where \(p_1, \dots , p_u\) are primes and \(\alpha_1, \dots , \alpha_u\) non zero integers.

If \(H \le G\) we have \[H \simeq \bigoplus_{i=1}^u \mathbb{Z}_{p_i^{\beta_i}}\] with \(0 \le \beta_1 \le \alpha_1, \dots, 0 \le \beta_u \le \alpha_u\). Then \[G/H \simeq \bigoplus_{i=1}^u \mathbb{Z}_{p_i^{\alpha_i-\beta_i}}\] which is a subgroup of \(G\).

*If \(G\) is not abelian, then \(G/H\) might not be isomorphic to a subgroup of \(G\).* Continue reading Is the quotient group of a finite group always isomorphic to a subgroup?