Applicable Group Problem Involving Modular Arithmetic

Question

Question: Show that $(5, 15, 25, 35)$ is a group under multiplication modulo 40.

So I first decided to make a Cayley Table that looks like this:

I apologize that I do not know how to make a Cayley Table in MathJax.

So I know it is closed, and associative because of multiplication inheritance.

I am just confused about the identity element. I could use some descriptive advice about how you find it. I know once you find that, you can use the table to find inverses of each.

Answer 1

Your identity element is $25$ because in mod $40$ you have

$$ 25\times 5=125 \equiv 5$$

$$ 25\times 15=375\equiv 15$$

$$25\times 25=625\equiv 25$$ $$25\times 35=875\equiv 35$$

It is interesting to see that for this group every element is its own inverse.

Answer 2

From the Cayley table we recognize the identity element ($25$, in red) because of the identities of the underlying set $\{5,15,25,35\}$ in the line and column labelled by it (in blue): \begin{array}{c|cccc} \times_{40} & 5 & 15 & \color{red}{25} & 35\\ \hline 5 & 25 & 35 & \color{blue}{5} & 15\\ 15 & 35 & 25 & \color{blue}{15} & 5\\ \color{red}{25} & \color{blue}{5} & \color{blue}{15} & \color{blue}{25} & \color{blue}{35}\\ 35 & 15 & 5 & \color{blue}{35} & 25\\ \end{array}

Associativity is well known to be verified since the operation is a restriction to a closed subset of $\mathbb{Z}_{40}$ of an operation which is associative on $\mathbb{Z}_{40}$.

But now hold on a moment.

$\mathbb{Z}_{40}$ is not a group under multiplication, it is only a (commutative) monoid. So - this is very important - claiming that $\{5,15,25,35\}$ is a closed subset of $\mathbb{Z}_{40}$ under multiplication does not imply that it is a group. It is a sub-semigroup of $\mathbb{Z}_{40}$ hence it is a commutative semigroup; it has identity element (not even the same as $\mathbb{Z}_{40}$) hence it is a commutative monoid. You had only proved that your structure is a commutative monoid, not a group.

How to prove that it is a group?

You had missed to say that all the lines and columns of our Cayley table are bijections of $\{5,15,25,35\}$, so that is the Cayley table of a quasigroup too. This wasn't trivial: not even $\mathbb{Z}_n\backslash\{0\}$ is a group under multiplication (and the lines and columns of its Cayley table are in general not bijections of $\mathbb{Z}_n\backslash\{0\}$) if $n$ isn't prime, and $40$ isn't prime.

Only now, a structure which is a commutative monoid and a quasigroup is a commutative (or Abelian) group.

Finally, the property "every element is its own inverse" stated in the answer by Mohammad may be reworded saying that the structure is unipotent (all the elements of the diagonal of the Cayley table are equals to the identity element).

So, our finite group of order $4$ is an unipotent Abelian group.

See also this other answer to a very similar question.