$|R|=30$ and $|I|=10$ then $I$ is a maximal ideal

Question

How shall I check this:
Suppose that $R$ is a commutative ring and $|R|=30.$ If $I$ is an ideal of $R$ and $|I|=10$ show that $I$ is a maximal ideal.
Please give some hint how to go with solution...

Answer 1

1) Any ideal which is a maximal additive subgroup is also a maximal ideal. (This is trivial.)

2) Any subgroup of a finite group whose index is a prime number is a maximal subgroup. (Follows from Langrage's Theorem)

Now conclude.

Answer 2

If another additive subgroup of $R$ sits between $I$ and $R$, its order would be a divisor of $30$ divisible by $10$. What could it be?

There are only two possibilities, showing that nothing lies strictly between.

You can easily see how this generalizes to any subgroup of order $n$ in a group of order $np$, $p$ prime (and hence you can pull the same trick for an ideal of size $n$ in a ring of size $np$.)