To which group is the subgroup isomorphic?

Question

We have the following subgroup of the dihedral group $D_6=\langle \sigma , \tau \mid \sigma^6=id=\tau^2, \tau\sigma=\sigma^5\tau\rangle$ $$\{\langle \sigma^2\rangle, \langle \sigma^2\rangle\sigma , \langle \sigma^2\rangle\tau , \langle \sigma^2\rangle\sigma \tau\}$$

This subgroup has $4$ elements, so it is isomorphic either to $\mathbb{Z}_4$ or to $\mathbb{Z}_2\times\mathbb{Z}_2$. To check to which of them it is isomorphic to, we have to check the order of each element, or not? Do all the elements of the above subgroup have the same order?

Also, an other subgroup is $$\{\langle \sigma^3\rangle, \langle \sigma^3\rangle\sigma , \langle \sigma^3\rangle\sigma^2, \langle \sigma^3\rangle\tau, \langle \sigma^3\rangle\sigma\tau, \langle \sigma^3\rangle\sigma^2\tau\}$$

This subgroup has $6$ elements, so it is isomorphic either to $\mathbb{Z}_6$ or to $D_3$. To check to which of them it is isomorphic to, what do we have to do?

Answer 1

For the first quotient,

$(\langle \sigma^2 \rangle \sigma)^2=\langle \sigma^2 \rangle \sigma^2=\langle \sigma^2 \rangle$ since $\sigma^2 \in \langle \sigma^2 \rangle$.

$(\langle \sigma^2 \rangle \tau)^2=\langle \sigma^2 \rangle \tau^2=\langle \sigma^2 \rangle \,id=\langle \sigma^2 \rangle$

$(\langle \sigma^2 \rangle \sigma \tau)^2=\langle \sigma^2 \rangle \sigma \tau \sigma \tau=\langle \sigma^2 \rangle \sigma \sigma^5 =\langle \sigma^2 \rangle \,id=\langle \sigma^2 \rangle$

So everything has order $2$ and so this quotient is $\mathbb{Z}_2 \times \mathbb{Z}_2$. The second quotient in a similar way is $D_3$.