Prove that if G is not equal to {e} and G has no proper non-trivial subgroup,then G is finite and o(G) is a prime number.?

Jun 21, 2018

If $G$ is infinite it has infinitely many subgroups, so it can't be infinite.

If $o \left(G\right)$ is not prime, there exists $g \setminus \in G$, whose order divides $o \left(G\right)$ that generates a non-trivial subgroup.

Explanation:

$G$ must be finite, because there's a theorem stating that

Finite number of subgroups $\setminus \implies$ finite group

So, using $\left(A \setminus \implies B\right) \setminus \implies \left(\setminus \neg B \setminus \implies \setminus \neg A\right)$

we deduce that an infinite group has an infinite number of subsets, and so $G$ must be finite, otherwise it would have subgroups and we are assuming it doesn't.

Now, let's prove that the order of $G$ must be prime: the order of every element $g \setminus \in G$ divides the order of the group. So, if $o \left(G\right)$ is not prime, it has proper divisors, say $k | o \left(G\right)$, and $k \setminus \ne 1$ and $k \setminus \ne o \left(G\right)$.

So, there exists an element $g \setminus \in G$ with $o \left(g\right) = k$, which would define a subgroup generated by powers of $g$.

Instead, if $o \left(G\right)$ is prime, the order of every $g \setminus \in G$ can either be $1$ or $o \left(G\right)$. If the order is $1$ the element is $e$, and the subgroup it generates is the trivial one.

If the order is $o \left(G\right)$, the subgroup generated is the whole group $G$. So, the only subgroups are the trivial ones.