Let σ= (1 2) (3 4 5) and τ= (1 2 3 4 5 6 ) be permutations in $S_6$.Which of the following statements are true? 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?
 A: (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. 
