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\headline{\vtop{\hbox to \hsize{\strut Math 2106 - Dr. Gupta\hfil Due Thursday
2022-04-14 at 11:59 pm}\hrule height .5pt}}
\centerline{\bigbf Homework 11 - Holden Rohrer}
\bigskip
\noindent{\bf Collaborators:} None
\section{Judson 5.4: 1b, 2c, 2p, 3a, 3c, 24}
\problem{1b.}
Write the following permutation in cycle notation:
$$\pmatrix{1&2&3&4&5\cr 2&4&1&5&3}.$$
\answer
This is $(1 2 4 5 3).$
\endanswer
\problem{2c.}
Compute $(1 4 3)(2 3)(2 4).$
\answer
$$\pmatrix{1&2&3&4\cr 2&3&1&4}$$
\endanswer
\problem{2p.}
Compute $[(1 2 3 5)(4 6 7)]^{-1}.$
\answer
$$\pmatrix{1&2&3&4&5&6&7\cr 5&1&2&7&3&4&6}$$
\endanswer
\problem{3a.}
Express the following permutation as a product of transpositions and
identify it as even or odd: $(1 4 3 5 6).$
\answer
$$(1 4 3 5 6) = (1 4)(4 3)(3 5)(5 6),$$
and since there are 4 transpositions, this is an even permutation.
\endanswer
\problem{3c.}
Express the following permutation as a product of transpositions and
identify it as even or odd: $(1 4 2 6)(1 4 2).$
\answer
$$(1 4 2 6)(1 4 2) = (1 4)(4 2)(2 6)(1 4)(4 2),$$
and since there are 5 transpositions, this is an odd permutation.
\problem{24.}
Show that a 3-cycle is an even permutation.
\answer
Let us have a 3-cycle $(a_1 a_2 a_3).$
This can be written as $(a_1 a_2)(a_2 a_3)$ because $a_2$ ends in the
position of $a_3,$ $a_1$ ends in the position of $a_2,$ and $a_3$ ends
in the position of $a_1.$
This is two transpositions, so this is an even permutation.
\endanswer
\section{Judson 6.5: 5d, 5b}
\problem{5d.}
List the left and right cosets of the subgroups of $A_4$ in $S_4.$
\answer
The left and right cosets are $\{A_4, (1\,2)A_4\}.$
\endanswer
\problem{5b.}
List the left and right cosets of the subgroups of $\langle 3\rangle$ in
$U(8).$
\answer
$\langle 3\rangle = \{{\rm id}, 3\},$ and since this group is abelian,
it has the same left and right cosets.
$5\langle 3\rangle = \{5, 7\},$ and this completes the partition of the
group.
\endanswer
\section{Problem not from the textbook}
\problem{1.}
Let $H$ be a subgroup of $G$ and suppose that $g_1,g_2\in G.$
Prove that $g_1H = g_2H$ if and only if $g_2\in g_1H.$
\answer
Let $H$ be a subgroup of $G$ and $g_1,g_2\in G.$
We will show that $g_1H = g_2H$ if and only if $g_2\in g_1H.$
$(\Rightarrow)$
Let $g_1H = g_2H.$
We will show that $g_2\in g_1H.$
$e\in H,$ so $g_2e = g_2\in g_2H,$ and by equality, $g_2\in g_1H.$
$(\Leftarrow)$
Let $g_2 \in g_1H.$
This means there is $h\in H$ such that $g_2 = g_1h.$
We then know that $g_2H = g_1hH = g_1H$ because $hH = H$ by group
closure.
\endanswer
\bye
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