Dummit And Foote Solutions Chapter 4 Overleaf High Quality -

\subsection*Problem S4.2 \textitLet $G$ be a cyclic group of order $n$. Prove that for each divisor $d$ of $n$, there exists exactly one subgroup of order $d$.

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

\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 $\langle 12/d \rangle$. \subsection*Problem S4

\subsection*Exercise 4.1.1 \textitProve that every cyclic group is abelian. Dummit And Foote Solutions Chapter 4 Overleaf High Quality

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

\beginsolution Let $|G| = p^2$. The center $Z(G)$ is nontrivial by the class equation (since $|G| = |Z(G)| + \sum |G:C_G(g_i)|$, each term divisible by $p$). So $|Z(G)| = p$ or $p^2$.