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\title{\sc Math 31 -- Homework 5 \framebox{Solutions!}}
\author{due \textbf{Friday, August 4}}
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\textbf{Instructions:} This assignment is due at the \emph{beginning} of class. Staple your work together (do not just fold over the corner). Please write the questions in the correct order. If I cannot read your handwriting, you won't receive full credit.
\begin{enumerate}
\item (10.8) Let $G$ be a group of order $p^2$, where $p$ is a prime. Show that $G$ must have a subgroup of order $p$.
\textbf{Proof:} Let $G$ be a group of order $p^2$ for a prime $p$. Let $x$ be a non-identity element. By Theorem 10.4, the possible orders of $x$ are only $p$ and $p^2$. If $o(x) = p$, then $\langle x \rangle$ is a subgroup of order $p$. So, in this case we are done. Otherwise $o(x) = p^2$.
If $o(x) = p^2$ then we know that $\langle x \rangle = G$, i.e., $G$ is cyclic with $x$ as a generator. It must follow then that $o(x^p) = p$, and $\langle x^p \rangle$ is a subgroup of $G$ of order $p^2/p = p$. \hfill $\square$
\hrulefill
\item (10.21) Prove that every group of order $77$ has an element of order $7$ and an element of order $11$.
\textbf{Proof:} Let $G$ be a group of order $77$ and let $x$ be a non-identity element. As $77 = 7 \cdot 11$ and both $7$ and $11$ are prime, the only possible orders of $x$ are $7$, $11$, and $77$.
If $G$ has an element $g$ of order $77$, then $G = \langle g \rangle$, and by Theorems from Chapter 4 it follows that $o(g^7) = 11$ and $o(g^{11}) = 7$, proving the result. Hence, we can now assume that every non-identity element of $G$ has order $7$ or $11$.
Suppose now toward a contradiction that $G$ has no element of order $7$, so that all elements of $G$ have only orders $1$ and $11$. We will show this is not possible. Any two non-trivial proper subgroups of $G$ have order $11$, and if they are not the same subgroup then their intersection is only $\{e\}$. This implies we can write $G$ as the union of a list of subgroups
\[
G = H_1 \cup H_2 \cup \cdots \cup H_k,
\]
such that $|H_i| = 11$ for all $i$, and $H_i \cap H_j = \{e\}$ for all $i \neq j$. If this is the case, then the size of $G$ can be computed as
\[
|G| = 1 + 10k,
\]
where the $1$ counts the identity element and $10k$ counts the ten non-identity elements in each subgroup, which must be different from those in the other subgroups. But, $77$ does not have the form $1 + 10k$. So this is not possible. It thus follows that $G$ does have an element of order $7$.
To see that $G$ also has an element of order $11$, assume again by contradiction that it does not. The same line of reasoning leads to the statement that
\[
77 = 1 + 6k
\]
for some integer $k$, which is not possible. \hfill $\square$
\hrulefill
\item (11.29) Show that if $G/Z(G)$ is cyclic then $G$ is abelian.
\textbf{Proof:} Let $G$ be a group and suppose that $G/Z(G)$ is cyclic. The elements of $G/Z(G)$ are right cosets of the form $Z(G)g$, for $g \in G$. If $G/Z(G)$ is cyclic, then there exists an element $Z(G)x \in G/Z(G)$ such that
\[
G/Z(G) = \langle Z(G)x \rangle.
\]
Note that by the definition of the operation on quotient groups,
\[
(Z(G)x)^i = Z(G)x^i,
\]
and so we can write that
\[
G/Z(G) = \{ Z(G)x^i : i \in \Z \}.
\]
To show that $G$ is abelian, we pick two arbitrary elements $a,b \in G$, and show that $ab = ba$. Consider the right cosets $Z(G)a$ and $Z(G)b$. By our earlier discussion, there exist $m$ and $n$ in $\Z$ such that
\[
Z(G)a = Z(G)x^m \qquad \text{ and } \qquad Z(G)b = Z(G)x^n.
\]
Therefore, we can write
\[
a = z_1x^m \qquad \text{ and } \qquad b = z_2x^n,
\]
where $z_1$ and $z_2$ are some elements of $Z(G)$.
Lastly,
\begin{align*}
ab &= (z_1x^m)(z_2x^n)\\
&= z_1z_2x^{m+n} \tag{because $z_2 \in Z(G)$}\\
&= z_2z_1x^{n+m}\tag{because $z_1,z_2 \in Z(G)$}\\
&= z_2z_1x^nx^m\\
&= (z_2x^n)(z_1x^m)\tag{because $z_1 \in Z(G)$}\\
&= ba.
\end{align*}
\hfill $\square$
\hrulefill
\item (11.16) Show that $(\Q,+)/(\Z,+)$ is an infinite group every element of which has finite order.
\textbf{Proof:} Let $G = (\Q,+)/(\Z,+)$. For ease of notation, we abbreviate $(\Q,+)$ and $(\Z,+)$ as $\Q$ and $\Z$. The elements of $G$ are right cosets of the form
\[
G = \left\{\Z + \frac{a}{b} \;:\; \frac{a}{b} \in \Q\right\}.
\]
So, we see that two right cosets $\Z + a/b$ and $\Z + c/d$ are the same if and only if $a/b$ and $c/d$ differ by an integer.
This allows us to list the distinct cosets in $\Q/Z$:
\[
G = \left\{\Z + \frac{a}{b} \;:\; \frac{a}{b} \in \Q \cap [0,1) \right\}.
\]
Since there are an infinite number of rational numbers between $0$ and $1$, the group $\Q/\Z$ is infinite.
What is the order of the coset $\Z + a/b$? Suppose that $a/b$ is written in reduced form. Then,
\[
\ell \cdot \left(\Z + \frac{a}{b}\right) = \Z + \left(\ell \cdot \frac{a}{b}\right),
\]
and this equals $\Z + 0$ if and only if $\ell \cdot \frac{a}{b}$ is an integer, which is true if and only if $\ell$ is a multiple of $b$. The smallest positive integer $\ell$ for which this is true is $\ell = b$. Therefore, $o(\Z + a/b) = b$, so long as $a/b$ is written in reduced form. \hfill $\square$
\hrulefill
\end{enumerate}
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