Problem 5.5.3 (Group $H \rtimes_\varphi Z_2$, where $H$ is abelian)

Question

In Example 1 following the proof of Proposition 11 prove that every element of $G - H$ has order $2$. Prove that $G$ is abelian if and only if $h^2 = 1$ for all $h \in H$.

Richard Ganaye · Accepted Answer

\begin{proof} 
In this example,
$$\varphi
\left\{
\begin{array}{ccl}
K  = \langle x angle =\{1,x\}& 	o & \mathrm{Aut}(H),\
k & \mapsto & \varphi_k
    \left\{
    \begin{array}{ccl}
    H & 	o & H\
    h &\mapsto &k\cdot h = 
    \begin{cases}
    h^{-1} & 	ext{if } k =x,\
    h & 	ext{if } k = 1.
    \end{cases}
    \end{array}
    ight. 
\end{array}
ight.
$$
is a homomorphism.


\bigskip
Let $(h, k )\in G - H$, i.e., $k 
e 1$, so $ k =x$. Then
\begin{align*}
(h,k)^2 &= (h\, (k\cdot h), k^2)\
&= (h \, (x \cdot h), k^2)\
&=(hh^{-1}, k^2)\
&= (1,1).
\end{align*}
so every element of $G - H$ has order $2$.

\bigskip
If $h^2 = 1$ for all $h \in H$, then $h^{-1} = h$, so $\varphi_k = \mathrm{id}_H$ for all $k \in K$, so $\varphi$ is the trivial homomorphism. By Theorem 11, $G \simeq H 	imes K$. Since $H$ and $K$ are abelian, $G$ is abelian.

Conversely, suppose that $G = H times_\varphi K$ is abelian. Then $K \unlhd H times_\varphi K$. By Theorem 11, $\varphi$ is the trivial homomorphism, thus $h^{-1} = x\cdot h = h$ for all $h \in H$, so $h^2 = 1$ for all $h \in H$.

In conclusion, $G$ is abelian if and only if $h^2 = 1$ for all $h \in H$.
\end{proof}

Problem 5.5.3 (Group $H \rtimes_\varphi Z_2$, where $H$ is abelian)

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Problem 5.5.3 (Group $H \rtimes_\varphi Z_2$, where $H$ is abelian)

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