# Is abelian group $G$ always isomorphic to H×(G/H)?

Let $G= \mathbb{Z}_4=\mathbb{Z}/4\mathbb{Z}$ and $H=\langle2\rangle$

Then $G= \mathbb{Z}_4 = \mathbb{Z}/4 \mathbb{Z} = \{0,1,2,3\}$.
We know $\mathbb{Z}/4\mathbb{Z}$ is cyclic.

Cayley Table of $G=\mathbb{Z}/4 \mathbb{Z}$ is:
$\begin{array}{|c|c|c|c|c|}\hline\hline\textbf{+} & \textbf{0} & \textbf{1} & \textbf{2} & \textbf{3}\\\hline \textbf{0} & 0 & 1 & 2 & 3 \\\hline \textbf{1} & 1 & 2 & 3 & 0 \\\hline \textbf{2} & 2& 3 & 0 & 1\\\hline\textbf{3} & 3 & 0 & 1 & 2 \\\hline \end{array}$   $\cong$   $\begin{array}{|c|c|c|c|c|}\hline\hline\bf{+} & \bf{I} & \bf{V_1} & \bf{V_2} & \bf{V_3}\\\hline \bf{I} & I & V_1 & V_2 & V_3 \\\hline \bf{V_1} & V_1 & V_2 & V_3 & I \\\hline \bf{V_2} & V_2& V_3 & I & V_1\\\hline\bf{V_3} & V_3 & I & V_1 & V_2 \\\hline \end{array}$

And $H=\langle2\rangle=\{0,2\}$.

Cayley Table of $H=\{0,2\}$ is:
$\begin{array}{|c|c|c|c|c|} \hline \hline \bf{+} &\bf{0}&\bf{2}\\ \hline \bf{0}&0&2\\\hline \bf{2} &2&0\\ \hline \end{array}$

Then, $G/H=\{\{0,2\},\{1,3\}\}$.
So, the group structures of $H, G/H, \mathbb{Z}_2$ are Isomorphic.
That is $H\cong G/H\cong\mathbb{Z}_2$.

And $H\times G/H=\{(0,\{0,2\}),(2,\{0,2\}),(0,\{1,3\}),(2,\{1,3\})\}$.
$(0,\{0,2\})$ is Identity in $H\times G/H$.

Cayley Table of $H\times G/H$ is:
$\begin{array}{|c|c|c|c|c|} \hline \hline \bf{+} &\bf{(0,\{0,2\})}&\bf{(2,\{0,2\})}&\bf{(0,\{1,3\})}&\bf{(2,\{1,3\})}\\ \hline \bf{(0,\{0,2\})} &(0,\{0,2\})&(2,\{0,2\})&(0,\{1,3\})&(2,\{1,3\})\\ \hline \bf{(2,\{0,2\})} &(2,\{0,2\})&(0,\{0,2\})&(2,\{1,3\})&(0,\{1,3\})\\ \hline \bf{(0,\{1,3\})} &(0,\{1,3\})&(2,\{1,3\})&(0,\{0,2\})&(2,\{0,2\})\\ \hline \bf{(2,\{1,3\})} &(2,\{1,3\})&(0,\{1,3\})&(2,\{0,2\})&(0,\{0,2\})\\ \hline \end{array}$

The Group structure of $H\times G/H$ is Isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$:
$\begin{array}{|c|c|c|c|c|} \hline \hline \bf{+} & \bf{I} & \bf{V_1} & \bf{V_2} & \bf{V_3} \\ \hline \bf{I} & I & V_1 & V_2 & V_3 \\ \hline \bf{V_1} & V_1 & I & V_3 & V_2 \\ \hline \bf{V_2} & V_2 & V_3 & I & V_1 \\ \hline \bf{V_3} & V_3 & V_2 & V_1 & I \\ \hline \end{array}$

Therefore, $H\times G/H \cong \mathbb{Z}_2\times\mathbb{Z}_2$ is not isomorphic to $\mathbb{Z}_4=\mathbb{Z}/4\mathbb{Z}$.