Dummit And Foote Solutions Chapter 4 Overleaf High | Quality

\beginsolution Groups of order 8: abelian: $\Z/8\Z$, $\Z/4\Z \times \Z/2\Z$, $\Z/2\Z \times \Z/2\Z \times \Z/2\Z$. Non-abelian: $D_8$ (dihedral), $Q_8$ (quaternion). So five groups total. \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

Divisors of 12: $1,2,3,4,6,12$. The subgroups are: \beginalign* &\langle 0 \rangle = \0\ \quad \text(order 1)\\ &\langle 6 \rangle = \0,6\ \quad \text(order 2)\\ &\langle 4 \rangle = \0,4,8\ \quad \text(order 3)\\ &\langle 3 \rangle = \0,3,6,9\ \quad \text(order 4)\\ &\langle 2 \rangle = \0,2,4,6,8,10\ \quad \text(order 6)\\ &\langle 1 \rangle = \Z_12 \quad \text(order 12) \endalign*

\newpage \section*Supplementary Problems for Chapter 4 Dummit And Foote Solutions Chapter 4 Overleaf High Quality

\subsection*Exercise 4.8.3 \textitShow that $\Inn(G) \cong G/Z(G)$.

\subsection*Exercise 4.4.7 \textitShow that $\Aut(\Z/8\Z) \cong \Z/2\Z \times \Z/2\Z$.

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. \beginsolution Groups of order 8: abelian: $\Z/8\Z$, $\Z/4\Z

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

If $|Z(G)| = p^2$, then $G$ is abelian. If $|Z(G)| = p$, then $G/Z(G)$ has order $p$, hence is cyclic. A well-known lemma states: if $G/Z(G)$ is cyclic, then $G$ is abelian. So $G$ is abelian in both cases. \endsolution

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

\beginsolution Let $G = \langle g \rangle$, $|G|=n$. For $d \mid n$, write $n = dk$. Then $\langle g^k \rangle$ has order $d$. Uniqueness: if $H \le G$, $|H|=d$, then $H = \langle g^m \rangle$ where $g^m$ has order $d$, so $n / \gcd(n,m) = d$, implying $\gcd(n,m) = k$. But $\langle g^m \rangle = \langle g^\gcd(n,m) \rangle = \langle g^k \rangle$. So unique. \endsolution

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