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Let $\sigma= (1 2) (3 4 5)$ and $\tau= (1 2 3 4 5 6 )$ be permutations in $S_6$, the group of permutations on six symbols. Which of the following statements are true?

a) The subgroups $\langle\sigma\rangle$ and $\langle\tau\rangle$ are isomorphic.

b) $\sigma$ and $\tau$ are conjugate in $S_6.$

c) $\langle\sigma\rangle \cap \langle\tau\rangle$ is the trivial group.

d) $\sigma$ and $\tau$ commute.

My attempt $o(\sigma)=$ l.c.m of ${2,3}=6$ and $o(\tau)=$ length of cycle=$6$ since $o(\sigma)=o(\tau)$ so $\langle\sigma\rangle$ and $\langle\tau\rangle$ are isomorphic to each other.

Which ones, among the next statements, are correct?

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    You have made a correct "attempt" for question a). Don't you think that you could easily answer to question d) as well ? – 2017-02-11
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    plz help me..i can not understand – 2017-02-11
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    I mean that in question d) you are asked to compose $\sigma \circ \tau$ and $\tau \circ \sigma$ and compare the results. Up to you, this is basic. If we do it for you, you will never learn... – 2017-02-11
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    I have calculated,σoτ ≠τoσ so they r not commutate – 2017-02-11

1 Answers 1

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(a) $\langle \sigma \rangle$ and $\langle \tau \rangle$ are isomorphic, since both elements are of order six, whence $$ \langle \sigma \rangle \cong \mathbf Z_6, \langle \tau \rangle \cong \mathbf Z_6, $$ and hence $$ \langle \sigma \rangle \cong \langle \tau \rangle. $$

(b) Two elements of $S_n$ are conjugate if only if they have the same cyclic structure, that is, their decompositions into product of disjoint cycles `look alike' (in particular, there is the same number of disjoint cycles in the corresponding decompositions). The standard proof is based on the fact that decomposition into a product of disjoint cycles in $S_n$ is unique up to order of factors, and on the useful formula \begin{equation*} \tag 1 \pi (i_1,\ldots,i_s) \pi^{-1}=(\pi(i_1),\ldots,\pi(i_s)) \end{equation*}

(c) Any power of $\sigma$ takes $1$ either to $1,$ or to $2.$ The only element of $\langle \tau \rangle$ which takes $1$ to $1$ is $\mathrm{id},$ and the only element which takes $1$ to $2$ is $\tau$ itself. But, on the other hand, any element of $\langle \sigma \rangle$ preserves $\{3,4,5\},$ which $\tau$ certainly doesn't. Thus $$ \langle \sigma \rangle \cap \langle \tau \rangle =\{\mathrm{id}\}. $$

(d) $\sigma$ and $\tau$ do not commute, for indeed $$ \tau \sigma \tau^{-1}=\tau\, (1,2)\,(3,4,5)\, \tau^{-1} =(\tau(1),\tau(2))\,(\tau(3),\tau(4),\tau(5))=(2,3)(4,5,6) \ne \sigma, $$ (use (1) for justification). Thus $$ \tau \sigma \tau^{-1} \ne \sigma \iff \tau \sigma \ne \sigma \tau, $$ as claimed.