$\{5,15,25,35\}$ is a group under multiplication mod $40$ 
Show that the set $\{5,15,25,35\}$ is a group under multiplication modulo $40$. What is the identity element of this group. Can you see any relationship between this group and $U(8)$?

I am very stuck on this question and I think my knowledge of abstract algebra is my liability at the moment. Right now I'm most confused by the fact that I learned: An element of a group of this nature must be relatively prime with the working mod.
Obviously here this is not the case, so I don't really know where to begin.
Do I need a Cayley Table?
Thanks.
 A: If you'd have {1,3,5,7} as a group under multiplication modulo 8 the canonic identity would be the 1. Since all element as well as the modulo factor are multiplied by 5 (twice for the product) the identity "shifts" to 25. 
The "relatively prime" statement works for the simple {1,3,5,7} case since they are relatively prime to 8 as well, but not for {5, 15, 25, 35} since they are not relatively prime to 40.
I don't know enough of the structure of unitary groups to see a relationship to your problem.
A: 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? 
A: Multiplication table proves this to be abelian group.
And it is isomorphic to U8
A: Remember that to prove a set $S$ under a binary operation $*$, written $\langle S, * \rangle$ is a group, you need to prove three properties of $\langle S, * \rangle$:


*

*For all $a,b,c \in S$, we have $(a * b) * c = a * (b*c)$, also known as associativity of $*$.

*There exists an $e \in S$ such that $\forall x \in S$, we have $e * x = x * e = x$. This is also known as the identity element $e$ for $*$.

*For each $a \in S$, there exists an $a^{-1} \in S$ such that $a * a^{-1} = a^{-1} * a = e$. This is known as the inverse $a^{-1}$ of $a$. Note that the inverse for each element is unique (you do not have to prove this).


So the proof for this is actually quite straightforward: for 1., you have to prove associativity of multiplication mod 40. Use the definition of modulo to help you. ($a \equiv b \mod c$ iff $c|(b-a)$). Then for 2. and 3., you just have to go through all the cases: four for 2. and four for 3.
EDIT: My post assumes that multiplication mod 40 is a valid binary operation. If you're not sure whether it is, then to prove that $*$ is a valid binary operation, you have to prove these two axioms:


*

*$*$ is well defined: exactly one element $s \in S$ is assigned to each possible ordered pair $(a,b) \in S \times S$. If you're having trouble with this, remember that a binary operation $*$ is formally a function $*: S \times S \to S$, and writing $a * b$ really means $*(a,b)$.

*$S$ is closed under $*$: For each ordered pair of elements of $S$, the element assigned to it is again in $S$.

A: You don't need to use cayley tables for this, you can prove it directly by noting that the elements are precisely those that can be written$5(2j+1)$ and using $\mod 40$ arithmetics.
You see immediately that it's closed under multiplication as $$5(2j+1) \times 5(2k+1) = 5( 2(10jk + 5j + 5k + 2) + 1)$$
It's obvious that multiplication is associative. Finding a candidate for the identity is to find a $j$ such that for all $k$
$$5( 2(10jk + 5j + 5k + 2) + 1) \equiv 5(2k+1) \mod 40$$
$$2(10jk + 5j + 5k + 2) + 1 \equiv 2k + 1 \mod 8$$
$$2(10jk + 5j + 5k + 2) \equiv 2k \mod 8$$
$$10jk + 5j + 5k + 2 \equiv k \mod 4$$
$$4\mid 10jk + 5j + 4k + 2$$
$$4\mid 2jk + j + 2$$
Now as this has to hold for all $k$ especially $0$ we have that $4\mid j+2$ which means that $j$ is even so $4\mid 2jk$ so we have that $j = 4n+2$ and since they are eqivalent mod $4$. We have the only possibility that $j=2$ and the corresponding element is $5(2j+1) = 25$. We see that this is indeed an identity.
For the existence of the inverse we use the same tecnique that for each $j$ there must be a $k$ such that:
$$5( 2(10jk + 5j + 5k + 2) + 1) \equiv 25 \mod 40$$
$$2(10jk + 5j + 5k + 2) + 1 \equiv5 \mod 8$$
$$2(10jk + 5j + 5k + 2)  \equiv 4 \mod 8$$
$$10jk + 5j + 5k + 2  \equiv 2 \mod 4$$
$$4 \mid 10jk + 5j + 5k$$
$$4 \mid 2jk + j + k$$
So we see that $2jk$ is even so $j+k$ must be even so we can write $j=k+2n$ for some $n$ and we get
$$4 \mid 2(k+2n)k + k+2n + k$$
$$4 \mid 2k^2+4nk + 2k +2n$$
$$2 \mid k^2 + k +n$$
$$2 \mid k(k+1) +n$$
We know that $k(k+1)$ is even so $n$ must be even that it's $2m$ for some $m$. So we have that the inverse is 
$$5(2k+1) = 5(2(j+4m) + 1) =  5(2j+9) = 5(2j+1)+40 \equiv 5(2j+1)\mod 40$$
That is every element is it's own inverse.
A: I have shown in this answer to a more recent but very similar question that the set is a group under the given operation.
For a finite magma, the two set of properties


*

*associativity + identity element + inverses (as in the answer to this question by Newb)

*associativity + identity element + cancellation (as in mine linked above)


are equivalent and mean the magma is a group.

Re-labeling the group subject of this question:
\begin{array}{c|cccc}
\circ & 5 & 15 & 25 & 35\\
\hline
5 & 25 & 35 & 5 & 15\\
15 & 35 & 25 & 15 & 5\\
25 & 5 & 15 & 25 & 35\\
35 & 15 & 5 & 35 & 25\\
\end{array}
we can get:
\begin{array}{c|cccc}
\bullet & 5 & 7 & 1 & 3\\
\hline
5 & 1 & 3 & 5 & 7\\
7 & 3 & 1 & 7 & 5\\
1 & 5 & 7 & 1 & 3\\
3 & 7 & 5 & 3 & 1\\
\end{array}
This latter is very similar to the group of units of the ring $\mathbb{Z}/8\mathbb{Z}$, namely $\mathbb{U}(8)$:
\begin{array}{c|cccc}
* & 1 & 3 & 5 & 7\\
\hline
1 & 1 & 3 & 5 & 7\\
3 & 3 & 1 & 7 & 5\\
5 & 5 & 7 & 1 & 3\\
7 & 7 & 5 & 3 & 1\\
\end{array}
We can get one from the other just by a permutation of the labels of the rows and columns of the table. This means the two groups are isomorphic: let $A=\{5,15,25,35\}$ and $B=\{1,3,5,7\}$ non-ordered sets, a bijection $f$ exists from $(A,\circ)$ to $(B,*)$ such that
$$f(x\circ y)=f(x)*f(y)$$
for all $x,y\in A$.

So the Cayley table is not mandatory (see the other five good answers without it) but it is useful, when it is given or is not too long to write down.
