2 seemingly isomorphic groups Take the following two groups:
$G_1$
$$\begin{array}{c|c|c|c|c}
 \cdot &amp; e
 &amp; a &amp; b&amp; c\\\hline
 e &amp; e                                       &amp; a    &amp; b   &amp; c \\\hline
 a &amp;a                                      &amp; e  &amp; c&amp;  b\\\hline
 b &amp; b                                       &amp; c    &amp; e &amp; a \\\hline
 c &amp; c                                       &amp; b &amp; a &amp; e
\end{array}$$
$G_2$
$$\begin{array}{c|c|c|c|c}
 \cdot &amp; e
 &amp; a &amp; b&amp; c\\\hline
 e &amp; e                                       &amp; a    &amp; b   &amp; c \\\hline
 a &amp;a                                      &amp; e  &amp; c&amp;  b\\\hline
 b &amp; b                                       &amp; c    &amp; a&amp; e \\\hline
 c &amp; c                                       &amp; b &amp; e&amp;a
\end{array}$$
In $G_1$ there are 3 normal subgroups, $\{e,a\},\{e,b\},\{e,c\}$
Each one leading to isomorphicly equivalent factor groups.
$$G_1 \cong \mathbb{Z} / 2\mathbb{Z} \times  \mathbb{Z} / 2\mathbb{Z} $$
$G_2$ has one normal subgroup $\{e,a\}$ which leads to
$$G_2 \cong \mathbb{Z} / 2\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z} $$
Which seems to imply there is an isomorphism between them, but there clearly isn’t.
Where am I wrong?
In$G_2$ we have $c=b^{-1}$ and $c^2=a$ so that $b^{-2}=a=b^2$, and hence $b^4=e$, but $b^2=a\neq e$. Hence $G_2$ has an element of order $4$, so that $G_2\cong C_4$, and hence is not isomorphic to $C_2\times C_2\cong G_1$.
No reason to think the groups are isomorphic.
In fact, they have different &quot;identity-skeletons&quot;, so can't be.