I think all you need has been pointed here. You don't need to write down the associated Cayley table for the presented set and its operation, but for this one it leads you to get the answer graphically:
We can see that the operation is a binary one. Can you find the identity element? What about the inverses ones?
Let $\psi(d)$ denote the number of elements of order $d$ in $\mathbb Z_p^*$.
Note that if $a\in\mathbb Z_p^*$ has order $d$, then $a^r$ has order $d$ if and only if $\gcd(r,d)=1$.
Thus are precisely $\phi(d)$ elements of the form of $a^r$ which have order $d$.
It follows that $\phi(d)$ divides $\psi(d)$ for all $d$.
Using (1) we want to show that $\psi(d)$ is either $0$ or equal to $\phi(d)$ for all $d|(p-1)$.
Assume on the contrary that there is some $d$ dividing $p-1$ such that $\psi(d)>\phi(d)$.
Let $a$ be an element of order $d$.
Out of the $d$ elements $a,a^2,\ldots,a^d$, precisely $\phi(d)$ have order $d$.
By our assumption, there is an element $b$ of order $d$ which is not equal to any of the $a^i$'s.
But note that $b$ and each $a^i$ satisfies $x^d\equiv 1\pmod{p}$.
This means that the congruence $x^d\equiv 1\pmod{p}$ has more than $d$ solutions, contrary to (1).
So we have $\psi(d)=0$ or $\phi(d)$.
Now note that $\sum_{d|(p-1)}\psi(d)=p-1$.
This is because each element of $\mathbb Z_p^*$ has order $d$ for some $d|p-1$.
From our earlier inference we also have $\sum_{d|p-1}\psi(d)\leq \sum_{d|p-1}\phi(d)$.
The RHS is equal $p-1$ by (2).
Therefore we cannot have $\psi(d)=0$ for any $d|p-1$ and hence $\psi(p-1)>0$, proving there is an element of order $p-1$ (and hence precisely $\phi(p-1)$ such elements).
Best Answer
I think you may need an answer like this:
The group includes $5$, since this is a group under multiplication modulo $56$, it has $1,5,25,13,9,45$.
Secondly, the group contains $15$ and $15$ is not in those six numbers, so the group consist of $15$ times each of $1,5,25,13,9,45$, that is, $15,19,39,27,23,3$.
And for the question "why can there be such a 12-element group", it is because the multiplication under modulo $56$ is an abelian group, thus it can be written as a direct sum of cycles. Some cycle will be even, thus if we take half of that cycle and other cycles, there will be a $12$-element subgroups.
I hope this helps.