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\noindent
Math 541 \\
Problem Set 5
\begin{enumerate}


\item[2.4.1.] Recall that
\[ \langle H \rangle = \bigcap_{H \subseteq K \le G} K. \]
$H$ is contained in each intersectand, hence in the intersection, that is, $H \subseteq \langle H \rangle$.  But $H \subseteq H \le G$, so $H$ is one of the intersectands, so $\langle H \rangle \subseteq H$.


\item[2.4.3.] We will argue that the generators commute with each other, so the subgroup they generate is abelian.

Let $x, y \in H \cup Z(G)$.  If $x \in Z(G)$ or $y \in Z(G)$ then $xy = yx$ by definition of the center.  If neither is in $Z(G)$ then $x, y \in H$, so again $xy = yx$ since $H$ is abelian.  Thus any two elements of $H \cup Z(G)$ commute.

Observe that if $x \in H \cup Z(G)$ then $x^{-1} \in H \cup Z(G)$ since $H$ and $Z(G)$ are subgroups.  Now an arbitrary element of $\langle H, Z(G) \rangle = \overline{H \cup Z(G)}$ is of the form $x_1 \dotsb x_m$, where $x_1 \dotsb x_m \in H \cup Z(G)$.  Let $y_1 \dotsb y_n$ be another element.  Then the $x_i$'s commute with the $y_i$'s, so $x_1 \dotsb x_m \cdot y_1 \dotsb y_n = y_1 \dotsb y_n \cdot x_1 \dotsb x_m$.

For the counterexample, let $G$ be any nonabelian group and $H = 1$, so $C_G(H) = G$, which is not ablian.  For a more satisfying counterexample, take $G = D_{8}$ and $H = \{ 1, r^2 \}$.  Then $r \in C_G(H)$, since $r^2 r = r r^2$, and $s \in C_G(H)$, since $r^2 s = r s r^{-1} = s r^{-2} = s r^2$ (since $r^4 = 1$).  Thus $C_G(H) = G$, which is not abelian.


\item[2.4.7.]  First, we figure out what this subgroup looks like.  Let $x = (1\ 2)$ and $y = (1\ 3)(2\ 4)$.  Since $x^2 = y^2 = 1$, the elements of $\langle x, y \rangle$ are strings of $x$'s and $y$'s such as $yxy$ and $xyxyx$; that is, we don't need powers of $x$ and $y$ in our strings.  Moreover, $xy = (1\ 3\ 2\ 4)$, so $(xy)^4 = 1$, so $y = xyxyxyx$, so we can assume that our strings start with $x$.  These are the possible strings:
\begin{align*}
x &= (1\ 2) &
xyxyx &= (3\ 4) \\
xy &= (1\ 3\ 2\ 4) &
xyxyxy &= (1\ 4\ 3\ 2) \\
xyx &= (1\ 4)(2\ 3) &
xyxyxyx &= (1\ 3)(2\ 4) \\
xyxy &= (1\ 2)(3\ 4) &
xyxyxyxy &= 1.
\end{align*}
In particular, $|\langle x, y \rangle| = 8$.

Now define a map $\phi: D_8 \to S_4$ by $\phi(s) = x$ and $\phi(r) = xy$.  This really does determine a group homomorphism, since it respects the relations: $\phi(s)^2 = 1$, $\phi(r)^4 = 1$, and $\phi(r)\phi(s) = \phi(s)\phi(r)^{-1}$.  Since $\im \phi$ is a subgroup containing $x = \phi(r)$ and $y = \phi(sr)$, $\langle x, y \rangle \subseteq \im \phi$.  Since $|D_8| = 8$ and $|\langle x, y \rangle| = 8$, this is the whole image, and $\phi$ is a bijection onto its image, hence an isomorphism onto its image.


\item[2.5.1.]
The most general lattice is
\begin{diagram}
& & G \\
& & \dLine \\
& & \langle H, K \rangle \\
& \ldLine & & \rdLine \\
H & & & & K \\
& \rdLine & & \ldLine \\
& & H \cap K \\
& & \dLine \\
& & \phantom{.}1.
\end{diagram}
If $H \cap K = 1$ then we contract the edge from $H \cap K$ to 1:
\begin{diagram}
& & G & & & & & & & & G \\
& & \dLine & & & & & & & & \dLine \\
& & \langle H, K \rangle & & & & & & & & \langle H, K \rangle \\
& \ldLine & & \rdLine & & & & & & \ldLine & & \rdLine \\
H & & & & K & & \rightsquigarrow & & H & & & & K \\
& \rdLine & & \ldLine & & & & & & \rdLine & & \ldLine \\
& & H \cap K & & & & & & & & \phantom{.}1. \\
& & \dDots \\
& & 1
\end{diagram}
If $\langle H, K \rangle = G$ then we contract the edge from $G$ to $\langle H, K \rangle$:
\begin{diagram}
& & G \\
& & \dDots \\
& & \langle H, K \rangle & & & & & & & & G \\
& \ldLine & & \rdLine & & & & & & \ldLine & & \rdLine \\
H & & & & K & & \rightsquigarrow & & H & & & & K \\
& \rdLine & & \ldLine & & & & & & \rdLine & & \ldLine \\
& & H \cap K & & & & & & & & H \cap K \\
& & \dLine & & & & & & & & \dLine \\
& & 1 & & & & & & & & \phantom{.}1.
\end{diagram}
If $K \subseteq H$ then $H \cap K = K$ and $\langle H, K \rangle = H$, so we contract two edges:
\begin{diagram}
& & G \\
& & \dLine & & & & & & G \\
& & \langle H, K \rangle & & & & & & \dLine \\
& \ldDots & & \rdLine & & & & & H \\
H & & & & K & & \rightsquigarrow & & \dLine \\
& \rdLine & & \ldDots & & & & & K \\
& & H \cap K & & & & & & \dLine \\
& & \dLine & & & & & & \phantom{.}1. \\
& & 1
\end{diagram}
Similarly, if $H \subseteq K$,
\begin{diagram}
& & G \\
& & \dLine & & & & & & G \\
& & \langle H, K \rangle & & & & & & \dLine \\
& \ldLine & & \rdDots & & & & & K \\
H & & & & K & & \rightsquigarrow & & \dLine \\
& \rdDots & & \ldLine & & & & & H \\
& & H \cap K & & & & & & \dLine \\
& & \dLine & & & & & & \phantom{.}1. \\
& & 1
\end{diagram}
Any combination of these four contractions could occur, making $2^4 = 16$ possible lattices.  We will not draw them all, but for example, we could have $\langle H, K \rangle = G$ and $K \subseteq H$, so
\begin{diagram}
G = H \\
\dLine \\
K \\
\dLine \\
\phantom{.}1.
\end{diagram}


\item[2.5.9.]
\begin{enumerate}
\item[(b)]
\begin{diagram}
& & & & & & \mathbb Z/24\mathbb Z \\
& & & & & \ldLine & & \rdLine(2,3) \\
& & & & \langle 2 \rangle \\
& & & \ldLine & & \rdLine(2,3) & & & \langle 3 \rangle \\
& & \langle 4 \rangle & & & & & \ldLine \\
& \ldLine & & \rdLine(2,3) & & & \langle 6 \rangle \\
\langle 8 \rangle & & & & & \ldLine \\
& \rdLine(2,3) & & & \langle 12 \rangle \\
& & & \ldLine \\
& & 1
\end{diagram}
\end{enumerate}


\end{enumerate}
\end{document}