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}.

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