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\title{\sc Math 31 -- Homework 6 \framebox{Solutions!}}
\author{due Wednesday, August 9}
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\date{}
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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 (12.13) Let $\phi : G \to H$ be a homomorphism.
\begin{enumerate}
\item Show that if $H$ is abelian and $\phi$ is one-to-one, then $G$ is abelian.\\
\textbf{Proof:} Suppose $H$ is abelian and $\phi$ is one-to-one. Let $a,b \in G$ be arbitrary. We aim to show that $ab = ba$.
By the rules of homomorphisms,
\[
\phi(ab) = \phi(a)\phi(b)
\]
and
\[
\phi(ba) = \phi(b)\phi(a).
\]
As $H$ is abelian,
\[
\phi(a)\phi(b) = \phi(b)\phi(a)
\]
and so
\[
\phi(ab) = \phi(ba).
\]
The injectivity of $\phi$ implies that $ab = ba$. \hfill $\square$
\item Show that if $G$ is abelian and $\phi$ is onto, then $H$ is abelian.\\
\textbf{Proof:} Suppose that $G$ is abelian and $\phi$ is onto. Let $x,y \in H$ be arbitrary. We aim to show that $xy=yx$.
Since $\phi$ is onto, there exist $a,b \in G$ such that $\phi(a) = x$ and $\phi(b) = y$. Now,
\[
xy = \phi(a)\phi(b) = \phi(ab)
\]
and
\[
yx = \phi(b)\phi(a) = \phi(ba).
\]
As $G$ is abelian, $ab = ba$. Therefore,
\[
xy = \phi(ab) = \phi(ba) = yx.
\]
Therefore $G$ is abelian. \hfill $\square$
\item Show that if $\phi$ is an isomorphism then $G$ is abelian if and only if $H$ is.\\
\textbf{Proof:} Suppose $\phi$ is an isomorphism.
If $G$ is abelian, then part (b) shows that $H$ is abelian because isomorphisms must be surjective. If $H$ is abelian, there part (a) shows that $G$ is abelian because isomorphisms must be injective. \hfill $\square$
\end{enumerate}
\hrulefill
\item (12.21) Let $G$ be the group of nonzero complex numbers under multiplication and let $H$ be the subgroup of $GL(2,\R)$ consisting of all matrices of the form $\ds\left(\begin{array}{cc}a&b\\-b&a\end{array}\right)$, where not both $a$ and $b$ are $0$. Show that $G \cong H$.
\textbf{Solutions:} To prove that $G \cong H$, we will define a function and prove that it's an isomorphism.
We need a function whose input is a nonzero complex number $a + bi$ and whose output is a nonzero matrix of the form $\ds\left(\begin{array}{cc}a&b\\-b&a\end{array}\right)$. The question was phrased in such a way to hint at the map
\[
\phi(a + bi) = \ds\left(\begin{array}{cc}a&b\\-b&a\end{array}\right).
\]
The map makes sense because we have to start with \emph{nonzero} complex numbers, and therefore the outputs of $\phi$ really are matrices of the proper form, with $a$ and $b$ not both $0$.
First, we'll check that $\phi$ is a homomorphism:
\[
\phi((a+bi)(c + di)) = \phi((ac - bd) + (ad + bc)i) = \ds\left(\begin{array}{cc}ac - bd&ad + bc\\-(ad + bc)&ac-bd\end{array}\right)
\]
while
\[
\phi(a+bi)\phi(c+di) = \ds\left(\begin{array}{cc}a&b\\-b&a\end{array}\right)\ds\left(\begin{array}{cc}c&d\\-d&c\end{array}\right) = \ds\left(\begin{array}{cc}ac - bd&ad + bc\\-(ad + bc)&ac-bd\end{array}\right).
\]
Next, we'll check that $\phi$ is surjective. Let $\ds\left(\begin{array}{cc}x&y\\-y&x\end{array}\right)$ be a matrix where $x$ and $y$ are not both zero. Then, $\phi(x + yi) = \ds\left(\begin{array}{cc}x&y\\-y&x\end{array}\right)$, and moreover $x + yi \neq 0$, so $x + yi$ really is in the domain.
Lastly, we'll verify that $\phi$ is injective. Let $r + si$ and $u + vi$ be two nonzero complex numbers such that $\phi(r+si) = \phi(u+vi)$. Then,
\[
\ds\left(\begin{array}{cc}r&s\\-s&r\end{array}\right) = \ds\left(\begin{array}{cc}u&v\\-v&u\end{array}\right).
\]
It follows that $r = u$ and $s = v$ so that $r + si = u + vi$. This confirms that $\phi$ is injective.
Having proved that $\phi$ is a homomorphism, injective, and surjective, we can conclude that $\phi$ is an isomorphism. Therefore,
\[
G \cong H.
\]
\hrulefill
\item (13.5) Let $G$ be the group of all real-valued functions on the real line, under addition of functions. Let $H$ be the subset of $G$ consisting of all $f$ such that $f(0) = 0$.
\begin{enumerate}
\item Show that $H \norm G$.
\textbf{Solution:} In part (b), we will find a surjective homomorphism $\phi : G \to (\R,+)$ whose kernel is $H$, then apply the first isomorphism theorem to conclude that $G/H \cong (\R,+)$. In fact, that reasoning also guarantees to us that $H \norm G$, because all kernels are normal in their domains. So, we don't really need to prove that $H \norm G$ on its own first.
Even if we didn't realize this, the proof is easy: $G$ is abelian (because additional is abelian), from which it follows that \emph{all} subgroups are normal. \hfill $\square$
\item Show that $G/H \cong (\R,+)$.
\textbf{Solution:} As outlined in part (a), our strategy requires us to find a surjective homomorphism $\phi : G \to (\R,+)$ whose kernel is $H$. A little reverse-engineering will prove enlightening. We know that we need to map each function to a real number, and the kernel of the map will be those functions that map to $0$. Moreover, we \emph{want} the kernel to contains those functions which take the value $0$ at $0$.
Now, with this in mind, suppose $f \in G$. What should $\phi(f)$ be? It must be a real number, and it must be $0$ when $f(0) = 0$. So, define $\phi(f) = f(0)$. (In words, $\phi$ takes as input a function $f$, and gives as output the value of $f$ at $0$.)
Is this a homomorphism? Well,
\[
\phi(f+g) = (f+g)(0) = f(0) + g(0) = \phi(f) + \phi(g),
\]
proving that it is.
Is $\phi$ surjective? Let $r \in \R$ be arbitrary. We only need to find a single function $f \in G$ such that $\phi(f) = r$. Choosing, for example, the constant function $c_r$ which takes value $r$ at all points suffices: $\phi(c_r) = c_r(0) = r$.
What is the kernel of $\phi$? By the definition of the kernel
\begin{align*}
\ker(\phi) &= \{ f \in G : \phi(f) = 0\}\\
&= \{ f \in G : f(0) = 0\}\\
&= H.
\end{align*}
Hence, by the first isomorphism theorem, we can conclude that
\[
G/\ker(\phi) \cong (\R, +),
\]
and so
\[
G/H \cong (\R,+).
\]
\ \hfill $\square$
\end{enumerate}
\hrulefill
\item (13.12) Let $G$ be a group, let $K \norm G$, and let $H$ be a subgroup of $G$ such that $HK = G$ and $H \cap G = \{e\}$. Show that $G/K \cong H$. (Note: $HK$ means the subgroup $\{hk : h \in H, k \in K\}$. One can prove that if $H$ and $K$ are subgroups of $G$, \emph{and} if $K$ is normal in $G$, then $HK$ is a subgroup of $G$.)
\textbf{Solution:} I believe that this question has a typo in the textbook, which I didn't notice until it was pointed out to me on Tuesday. Thankfully, even with the typo it is a true statement, but much more trivial. (Nonetheless, this is still good practice! If your proof is long, then you didn't understand the question well enough to realize it is trivial.) I believe that the author meant ``$H \cap K = \{e\}$'' rather than ``$H \cap G = \{e\}$'', which is how I initially read the question.
Let's now prove the statement as written. Let $G$ be a group, let $K \norm G$, and let $H \leq G$ such that $HK = G$ and $H \cap G = \{e\}$. If $H$ is a subgroup of $G$ such that $H \cap G = \{e\}$, then $H$ is itself trivial, i.e., $H = \{e\}$. Now, the only way that $HK = G$ is if $K = G$.
This leads us to conclude that $G/K = G/G$, which is a rather silly group consisting of one element (a coset): $G/G = \{G\} = \{e_{G/G}\}$. On the other hand, $H$ is also a group of size one: $H = \{e_G\}$. All groups of size $1$ are isomorphic (as evidenced, if it's really necessary, by the map that sends $e_{G/G}$ to $e_G$, which, being on a domain and codomain of one element each, is by necessity a bijection and a homomorphism). Therefore, $G/K \cong H$. \hfill $\square$
\hrulefill
\end{enumerate}
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