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 \(G\).*

### The case of the quaternion group \(Q_8\)

Consider the quaternion group \[Q_8 =\langle -1,i,j,k \ | \ (-1)^2=1,i^2=j^2=k^2=ijk=-1\rangle\]

The center of \(Q_8\) is \(Z=\{-1,1\}\) as we have the relations:\[\left\{

\begin{array}{ccc}

ij=k & , & ji=-k \\

jk=i & , & kj=-i \\

ki=j & , & ij=-i \\

\end{array}

\right.\]

\(Q_8/Z\) has order four and all elements of \(Q_8/Z\) have order two as for \(x \in Q_8\), \(x^2 \in \{-1,1\}\). On the other hand the order four subgroups of \(Q_8\) are \(\langle i \rangle, \langle j \rangle, \langle k \rangle\) which are cyclic and isomorphic to \(\mathbb Z_4\).

Finally \(Q_8/Z\) cannot be isomorphic to a subgroup of \(G\).