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1

In part b) you will need the fact $(A+B)^T=A^T+B^T$ and $(3A)^T=3A^T$and the same idea solves the problem.


5

Every symmetry of the triangle permutes its vertices, so there is an embedding $D_3\to S_3$ already just from geometry. Then it suffices to check they have the same order.


2

Hint: $S_3=\langle(1,2,3),(1,2)\rangle$ and $D_3=\langle a,b\rangle$ set $\phi(a)=(1,2,3)$ and $\phi(b)=(1,2)$ i.e send generaters to generaters. Note: Choose $a$ from the ones of order $3$ and b from the ones of order $2$.


2

Method 1 The most direct way is to use a presentation for the groups. You know a general dihedral group of order $2n$ has a presentation: $$D_{n}=\langle a,b | a^2=b^n=e, aba=b^{-1}\rangle$$ Then just establish that with $n=3$, and by choosing $a=(12)$ (or any transposition) and $b=(123)$ (or $(321)$) that the corresponding elements of the symmetric ...


3

Let me attempt the hardest part first, which is to show that $F(S)$ is a subgroup. Let $f, g: S\to S$ be permutations with finite support then observe that $$ \mathrm{Fix}(g^{-1}\circ f) = \{ x \mid g(x) = f(x)\} \supseteq \{ x \mid g(x)= f(x) = x \} = \mathrm {Fix}(g) \cap \mathrm{Fix}(f), $$ taking complements: $$ \mathrm{supp}(g^{-1}\circ f) \subseteq ...


1

I assume you haven't seen Lagrange's theorem, as that makes this question much easier. As a result, we can hash through this the old-fashioned way. Since subgroups are closed under inverses, and the $3$-cycles, $(123),(321)$ are inverses of one another, any subgroup of order $5$ must have both of them and the identity (because otherwise it is missing more ...


0

If there is a subgroup of order $5$ of a group with more than $5$ members, then some member is not in that subgroup, and that member is in some left coset of the subgroup of order $5$, and that left coset has $5$ members. What does that imply?


1

The size of the conjugacy class of an element $g$ in a group $G$ is the index of the subgroup consisting of the elements commuting with $g$. In the symmetric group, it's easy to work out the size of the conjugacy class of $g$, since two elements are conjugate in the symmetric group if and only if they have the same cycle structure. So all you have to do for ...


5

Given $\tau \in S_{10}$, the condition that $\tau$ should commute with $\sigma$ is equivalent to $\tau \sigma \tau^{-1} = \sigma$. But $\tau \sigma \tau^{-1} = (\tau(1) \ \tau(3) \ \tau(5) \ \tau(7) \ \tau(9))$. The condition can therefore be rewritten as $(\tau(1) \ \tau(3) \ \tau(5) \ \tau(7) \ \tau(9)) = (1 \ 3 \ 5 \ 7 \ 9)$. So $\tau(2)$, $\tau(4)$, ...


1

Note that instead of computing the size of the centralizer $C_{S_9}(a)$ of $a$ directly you can compute the size of the conjugacy class $C_a$ of $a$, because $|C_a| |C_{S_9}(a)| = |S_9| = 9!$. But the conjugacy class of $a$ just consists of all the permutations of the form $(s_1,s_2)(s_3,s_4,s_5)(s_6,s_7,s_8,s_9)$ with $\{s_1,\dots,s_9\} = \{1,\dots, 9\}$, ...


3

Consider, just for testing, $b=(124)(35)(6789)$ and compute \begin{align} b\star a\star b^{-1} &= (124)(35)(6789)(16)(257)(3849)(142)(35)(6987)\\ &=(1659)(27)(384)\\ &=\bigl(b(1)\,b(6)\bigr)\, \bigl(b(2)\,b(5)\,b(7)\bigr)\, \bigl(b(3)\,b(8)\,b(4)\,b(9)\bigr) \end{align} This holds in general (and you should be able to prove it): if $b$ is ...


1

You need to define a composition law, not just a set, for this to be a group. For this to be a group (and a subgroup of $S_{\mathbb{R}})$ you probably wanted to define composition as $f_n,f_m \mapsto f_n \circ f_m$. Then $f_n, f_m \mapsto (x + m) + n = x + (m+n) = f_{m+n}$. This shows that $G \cong \mathbb{Z}^+$ so it must be cyclic. Alternatively, you ...


0

$x+m$ is not an element of $G$. Perhaps what you meant to say is that $f_m$ is a generator for any $m$. But that is not true either. For example, how can you get $f_1$ by composing $f_2$ with itself?


1

If $n\ge 5$,for all $\tau=(i,j)(r,s)(i\not=j,r\not=s)$, we prove that this is in the subgroup generated by all $A_{n-1}^i$. This is because if $i=r,j=s$, $\tau=1$ is trivial, if $j=r,i\not=s$, $\tau=(jsi)=(i_1,i_2)(j,s)(s,i)(i_1,i_2)$, here $(i_1,i_2)(j,s)\in A^i_{n-1}$ and $(s,i)(i_1,i_2)\in A^j_{n-1}$($i_1,i_2$ are distinct and not the same as $i,j,s$) . ...


4

Hints: Write the permutation as a composite of disjoint cycles, some of which could have infinite length. It is then enough to solve the problem for the individual cycles. Any cycle of length greater than $2$ is a composite of two elements of order $2$. For example, $(1,2,3,4,5) = f \circ g$ with $f=(2,5)(3,4)$, $g=(1,5)(2,4)$. This also applies to cycles ...


1

We have $d=am+bn$ where $a$ and $b$ are integers. So, $f^d=f^{am}f^{bn}=\left(f^m\right)^a\left(f^n\right)^b=1$ because $f^n$ and $f^m$ are 1.


3

Apply the Bezout identity: $$ \exists x,y\in\mathbb Z, xm+yn=d $$ $$ f^d=f^{xm+yn}=(f^m)^x(f^n)^y=id $$



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