Exercise 2.4.3 ($\langle H, Z(G) \rangle$ is abelian)

Question

Prove that if $H$ is an abelian subgroup of a group $G$ then $\langle H, Z(G) \rangle$ is abelian. Give an explicit example of an abelian subgroup $H$ of a group $G$ such that $\langle H, C_G(H)\rangle$ is not abelian.

Richard Ganaye · Accepted Answer

\begin{proof} We write $Z = Z(G)$.

Since $H$ is abelian, $H \subseteq C_G(H)$, and $Z = C_G(G) \subseteq C_G(H)$, therefore $H \cup Z \subseteq C_G(H)$. Since $\langle H, Z\rangle$ is the smallest subgroup of $G$ containing $H \cup Z$, we obtain
\begin{align}
\langle H, Z\rangle \subseteq C_G(H).
\end{align}
Let $a \in \langle H, Z\rangle$. Then $a$ commutes with every element of $H$ by (1), and by definition of $Z$, $a$ commutes with every element of $Z$.

By Proposition 9, every element $b \in \langle H, Z\rangle$ is of the form $b = a_1^{\varepsilon_1}a_2^{\varepsilon_2}\cdots a_n^{\varepsilon_n}$ ($\varepsilon_i = \pm 1,\ a_i \in H \cup Z$). Since $a$ commutes with every $a_i$ (and consequently with $a_i^{-1}$), $a$ commutes with $b$. This shows that $ab = ba$ for every $a \in \langle H, Z\rangle$ and every $b \in \langle H, Z\rangle$.

Therefore $\langle H, Z(G) \rangle$ is abelian.

\bigskip
Consider the group $G = D_8$, and $H = \{e,r^2\}$. Then $H = Z(G)$ is the center of $G$, so $H$ is abelian, and $C_G(H) = G = D_8$ (see Ex. 2.2.4). Then $\langle H, C_G(H)\rangle = D_8$ is not abelian.
\end{proof}

Exercise 2.4.3 ($\langle H, Z(G) \rangle$ is abelian)

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Exercise 2.4.3 ($\langle H, Z(G) \rangle$ is abelian)

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