2 seemingly isomorphic groups

Question

Take the following two groups:
$G_1$ $$\begin{array}{c|c|c|c|c} \cdot & e & a & b& c\\\hline e & e & a & b & c \\\hline a &a & e & c& b\\\hline b & b & c & e & a \\\hline c & c & b & a & e \end{array}$$ $G_2$ $$\begin{array}{c|c|c|c|c} \cdot & e & a & b& c\\\hline e & e & a & b & c \\\hline a &a & e & c& b\\\hline b & b & c & a& e \\\hline c & c & b & e&a \end{array}$$ 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?

Answer 1

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 $C_2\times C_2\cong G_1$.