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2

This is incorrect, as Alastair's counterexample shows. Two related statements that are correct are: If $\sigma$ is an $n$-cycle (more generally, an $m$-cycle for $m | n$) then $\sigma^n = \iota$. For any finite group $G$, any element $g \in G$ satisfies $g^{\# G} = \iota$, and so for $\sigma \in S_n$, $\sigma^{n!} = 1$.

0

you rememmber wrong, consider $S_4$ and consider $\sigma =(1,2,3)$, then $\sigma^4=(1,2,3)=\sigma$. I think, what your professor said was if $\sigma$ a $n$-cycle, then $\sigma^4=e$

0

For an $n\times n$ matrix $A$ and a permutation $\sigma$, let $\sigma A$ denote the matrix obtained by using $\sigma$ to permute the rows of $A$. Then $A_{\sigma}=\sigma I$ and we have to show that $A_{\sigma}A_{\tau}% =A_{\sigma\tau}$. Explicitly, if $A=(a_{ij})$, then $\sigma A=(a_{ij}^{\prime})$ with $a_{ij}^{\prime}=a_{\sigma^{-1}(i)j}$. Use this to ...

1

If $a=b$ the question is the same as asking "Do every group have a maximal subgroup?" And some groups (like $\mathbb{Q}$) do not have this property.

3

$a,b\in cH$ iff $aH=cH$ and $a^{-1}b\in H$. So the proposition is true iff every element $g=a^{-1}b$ of $G$ is contained in a maximal subgroup of $G$. If $G$ is finite, that's true iff $G$ is not cyclic.

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.

6

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.

3

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 $2$.

3

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 ...

2

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 ...

1

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?

2

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 ...

6

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 ...

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