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\title\textbfDummit \& Foote \textitAbstract Algebra \\ Chapter 4 Solutions \authorYour Name \date\today

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\subsection*Exercise 4.5.9 \textitLet $G$ be a finite group and let $H$ be a subgroup of $G$ with $

Check powers of $r$: $r$ does not commute with $s$ since $srs = r^-1 \ne r$ unless $r^2=1$, but $r^2$ has order 2. Compute $r^2 s = s r^-2 = s r^2$ (since $r^-2=r^2$), so $r^2$ commutes with $s$. Also $r^2$ commutes with $r$, thus with all elements. $r$ and $r^3$ are not central. $s$ is not central (doesn’t commute with $r$). Similarly $rs$ not central.

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\beginsolution Let $[G:H] = 2$, so $H$ has exactly two left cosets: $H$ and $gH$ for any $g \notin H$. Similarly, the right cosets are $H$ and $Hg$. For any $g \notin H$, we have $gH = G \setminus H = Hg$. Thus left and right cosets coincide, so $H \trianglelefteq G$. \endsolution

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Hence $Z(D_8) = \1, r^2\ \cong \Z/2\Z$. \endsolution

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\beginsolution $D_8 = \langle r, s \mid r^4 = s^2 = 1, srs = r^-1 \rangle$. The center $Z(D_8)$ consists of elements commuting with all group elements.

\subsection*Exercise 4.1.1 \textitProve that every cyclic group is abelian.

\beginsolution Let $|H| = n$ and suppose $H$ is the only subgroup of $G$ with order $n$. For any $g \in G$, consider $gHg^-1$. Conjugation is an automorphism of $G$, so $|gHg^-1| = |H| = n$. Thus $gHg^-1$ is also a subgroup of $G$ of order $n$. By uniqueness, $gHg^-1 = H$ for all $g \in G$. Hence $H \trianglelefteq G$. \endsolution

\beginsolution Recall: \beginitemize \item Centralizer: $C_G(H) = \ g \in G \mid gh = hg \ \forall h \in H \$. \item Normalizer: $N_G(H) = \ g \in G \mid gHg^-1 = H \$. \enditemize If $g \in C_G(H)$, then for all $h \in H$, $ghg^-1 = h \in H$, so $gHg^-1 = H$. Hence $g \in N_G(H)$. Therefore $C_G(H) \subseteq N_G(H)$. Both are subgroups of $G$, so $C_G(H) \le N_G(H)$. \endsolution