Subgroups of a symmetric group

Got two questions on this that I can't seem to get my head around, then again maybe doing this with a headache isn't the best of scenarios

First one is below. How do I go about getting the elements for part (c) ??

It's just a cyclic subgroup, so keep calculating powers of $\alpha$ till you reach $\alpha^n = e$ for some (small) $n$. Then your subgroup is $\{e,\alpha,\alpha^2 ,\ldots, \alpha^{n-1}\}$.

Thanks, got it now

What about part (d) of this? How can I go about finding a subgroup here??

What was your answer to (c)?

I said that there is indeed a bijection $\phi : G \rightarrow \mathbb{Z}_8$ defined by the map below but I'm unsure if this can help in part (d) which I'm currently trying to follow from Cayley Th. proof.

A bijection isn't immediately an isomorphism (and typically won't be). In any case finding an explicit isomorphism is probably unnecessary here. It would be enough to find an order 8 element in G. Did you consider the elements' orders?

I did not consider the orders. I don't think there is an order 8 element in G otherwise I've done part (b) wrong...

ord(1)= 1
ord(2)=ord(7)=ord(8)=ord(13)= 4
ord(4)=ord(11)=ord(13)= 2

and I can see the orders are different for $\mathbb{Z}_8$.

Correct - so you've shown G isn't isomorphic to Z_8. Can you tell what order 8 group it is isomorphic to?

No I cannot unfortunately

No I cannot unfortunately

Also, I have followed Cayley's theorem, and so is it enough to say that if I use the mult. table's each row as a permutation $\sigma_i \in S_8$, then $\{ \sigma_1, \sigma_2, ..., \sigma_8 \}=G' \subset S_8$ is isomorphic to G ??

I have to say that given the explicit mention of Cayley's theorem before I'd have assumed this is what they're actually after, although I agree with your point that in many ways it's not the most informative way of looking at things.

I do think that it's good to do an explicit "build an isomorphism using the methodology in Cayley's theorem" once (or even twice!). So I'm partly thinking this may be that one time...