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\lhead{MATH 521--01}
\chead{Problem Set \#6\answer{ solutions}}
\rhead{}
\lfoot{}
\cfoot{\answer{Page \thepage\ of \pageref{LastPage}}}
\rfoot{due Thurssday, November 21, 2014}
\begin{document}
\begin{enumerate}
\item \question{Determine all of the homomorphisms from $Z_{20}$ to
itself. For each homomorphism, determine its kernel.}
\answer{Since $Z_{20}$ is cyclic, a homomorphism is uniquely
determined by the image of a generator (for simplicity, we'd
consider the image of $1$). There are thus 20 homomorphisms: we
could denote them $\varphi_0,\varphi_1,\ldots,\varphi_19$ according to the
definition $\varphi_k(1)=k$, which would induce the general rule
$\varphi_k(x)=kx$. The kernels will be as follows:
\begin{align*}
\ker\varphi_0&=Z_{20}=\{0,1,\ldots,19\}\\
\ker\varphi_1=\ker\varphi_3=\ker\varphi_7=\ker\varphi_9=\ker\varphi_{11}=\ker\varphi_{13}=\ker\varphi_{17}=\ker\varphi_{19}&=\{0\}\\
\ker\varphi_2=\ker\varphi_6=\ker\varphi_{10}=\ker\varphi_{14}=\ker\varphi_{18}&=\langle 10\rangle=\{0,10\}\\
\ker\varphi_4=\ker\varphi_8=\ker\varphi_{12}=\ker\varphi_{16}&=\langle 5\rangle=\{0,5,10,15\}\\
\ker\varphi_5=\ker\varphi_{15}&=\langle 4\rangle=\{0,4,8,12,16\}\\
\ker\varphi_{10}&=\langle 2\rangle=\{0,2,\ldots,18\}\\
\end{align*}
}
\item \question{Prove that if $H\trianglelefteq G$ and
$K\trianglelefteq G$ and $H\cap K=\{e\}$, then $G$ is isomorphic
to a subgroup of $G/H\oplus G/K$.}
\answer{We shall build an injective homomorphism $\varphi$ from $G$
into $G/H\oplus G/K$; since this is an injection, it can be made
bijective by limiting its domain, and will then become an
isomorphism, so that $\varphi(G)$ is a subgroup of $G/H\oplus G/K$
which is isomorphic to $G$.
Our homomorphism will be defined very na\"\i vely: let
$\varphi(g)=(gH,gK)$. It is abundantly clear by the way
multiplications of cosets of normal groups and multiplications in
direct products are calculated that this will in fact be a
homomorphism:
\[\varphi(g_1)\varphi(g_2)=(g_1H,g_1K)(g_2H,g_2K)=(g_1Hg_2H,g_1Kg_2K)=\left((g_1g_2)H,(g_1g_2)K\right)=\varphi(g_1g_2)\]
so now all that remains is to prove that this homomorphism is
injective. Suppose $\varphi(g)=\varphi(g')$. Thus
$(gH,gK)=(g'H,g'K)$, so $gH=g'H$ and $gK=g'K$. We may thus
conclude, based on our knowledge of when cosets coincide, that
$g^{-1}g'\in H$ and $g^{-1}g'\in K$. However, the only thing which
is both an element of $H$ and or $K$ is $e$, so $g^{-1}g'=e$ and
thus $g=g'$.}
\item \question{Prove that if $G$ and $H$ are finite groups and $|G|$
and $|H|$ are relatively prime, for any homomorphism
$\varphi:G\rightarrow H$, $\ker\varphi=G$.}
\answer{We know from the First Isomorphism Theorem (or even some
simpler results) that $|\varphi(G)|=\frac{|G|}{|ker\varphi|}$, so
$|\varphi(G)|$ is a divisor of $|G|$. Since $\varphi(G)\leq H$,
Lagrange's Theorem also requires that $|\varphi(G)|$ be a divisor
of $|H|$. However, since $|G|$ and $|H|$ are relatively prime,
their only common divisor is $1$, so $|\varphi(G)|=1$, and thus
the above equality derived from the First Isomorphism Theorem
becomes $1=\frac{|G|}{|\ker\varphi|}$, and so
$|\ker\varphi|=|G|$. Since $\ker\varphi\subseteq G$ and the two
sets have the same finite size, $\ker\varphi=G$.}
\item \question{Determine, with proof, how many nonisomorphic Abelian
groups there are of order 360.}
\answer{Note that $360=2^33^25^1$. We may say with absolute
certainty, from the Fundamental Theorem of Abelian Groups, that a
group $G$ of order 360 can be written as a direct product of three
groups of order $2^3$, $3^2$, and $5$. The third of these is
uniquely determined to be $Z_5$; the second could be either $Z_9$
or $Z_3\oplus Z_3$; and the first is one of $Z_8$, $Z_4\oplus
Z_2$, or $Z_2\oplus Z_2\oplus Z_2$. Thus, there are in fact 6
different Abelian groups of order 360.}
\item \question{You are told that a group $G$ is an Abelian group of
order 16 and that there are elements $a$ and $b$ of $G$ which both
have order 4, and for which $a^2\neq b^2$. Which isomorphism
classes could contain this group?}
\answer{There are 5 partitions of the number 4, which correspond to
5 possible Abelian groups of order 16 by the Fundamental Theorem
of Abelian Groups. So even without the condition on $a$ and $b$,
we can limit the candidates for group $G$ to the following four:
$Z_{16}$, $Z_8\oplus Z_2$, $Z_4\oplus Z_4$, $Z_4\oplus Z_2\oplus
Z_2$, and $Z_2\oplus Z_2\oplus Z_2\oplus Z_2$. We can consider
each of these individually.
\textbf{Case I: $G=Z_2\oplus Z_2\oplus Z_2\oplus Z_2$.} This group
has no elements of order 4 and thus no such $a$ and $b$ exist.
\textbf{Case II: $G=Z_8\oplus Z_2$.} $(x,y)$ has order 4 only if
at least one of $x$ and $y$ has order 4. No element of $Z_2$ has
order 4, and 2 and 6 are the only elements of $Z_8$ of order
4. Thus $a$ and $b$ must be of the form $(2,x)$ and/or
$(6,y)$. However $(2,x)^2=(4,0)$ and $(6,y)^2=(4,0)$, so $a^2$ and
$b^2$ cannot be unequal.
\textbf{Case III: $G=Z_4\oplus Z_4$.} The elements $a=(1,0)$ and
$b=(0,1)$ satisfy the given conditions on $a$ and $b$.
\textbf{Case IV: $G=Z_4\oplus Z_2\oplus Z_2$.} $(x,y,z)$ has order 4 only if
at least one of $x$, $y$, and $z$ has order 4. No element of $Z_2$ has
order 4, and 1 and 3 are the only elements of $Z_4$ of order
4. Thus $a$ and $b$ must be of the form $(1,y,z)$ and/or
$(3,y',z')$. However $(1,y,z)^2=(2,0)$ and $(3,y',z')^2=(2,0)$, so $a^2$ and
$b^2$ cannot be unequal.
\textbf{Case V: $G=Z_{16}$.} This group has exactly two elements
of order 4: 4 and 12, but $2\cdot 4=2\cdot 12$, so this does not
satisfy the necessary conditions for $a$ and $b$.
Thus we see that $G$ must be isomorphic to $Z_4\oplus Z_4$.
}
\end{enumerate}
\end{document}