Foote Solutions Chapter 4 Overleaf High Quality — Dummit And

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

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

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\subsection*Exercise 4.6.11 \textitFind the center of $D_8$ (the dihedral group of order 8). Dummit And Foote Solutions Chapter 4 Overleaf High Quality

\section*Chapter 4: Cyclic Groups and Properties of Subgroups \addcontentslinetocsectionChapter 4: Cyclic Groups

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

\beginsolution Define $\phi: G \to \Aut(G)$ by $\phi(g) = \sigma_g$ where $\sigma_g(x) = gxg^-1$. The image is $\Inn(G)$. Kernel: $\phi(g) = \textid_G$ iff $gxg^-1=x$ for all $x\in G$ iff $g \in Z(G)$. By the first isomorphism theorem, \[ G / Z(G) \cong \Inn(G). \] \endsolution \subsection*Exercise 4

\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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\beginsolution $\Z_12 = \0,1,2,\dots,11\$ under addition modulo 12. By the fundamental theorem of cyclic groups, for each positive divisor $d$ of 12, there is exactly one subgroup of order $d$, namely $\langle 12/d \rangle$. Kernel: $\phi(g) = \textid_G$ iff $gxg^-1=x$ for all

\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 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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