# If $G$ has no nontrivial subgroups, show that $G$ must be finite of prime order [duplicate]

We know the following fact from gorup theory: If $G$ is a group of prime order then it has no nontrivial subgroups.

Lets try to prove the converse statement: If $G$ has no nontrivial subgroups, show that $G$ must be finite of prime order.

Proof: Suppose by contradiction: If $G$ has no nontrivial subgroups $\Rightarrow$ $G$ is infiinite or $|G|\neq p$.

The case when $G$ is infinite we can rule out using that topic.

Suppose that $|G|=n$ where $n$ is composite $\Rightarrow$ $n=pm$ where $p$ - prime and $m\geqslant 2$. Let $a\in G$ such $a\neq e$ then $a^{pm}=e$.

1. If $a^p\neq e$ then considering the cyclic group of G, namely $H=\left \langle a^p\right \rangle$ $\Rightarrow$ $1<|H|\leqslant m<pm$ this is a contradiciton.

2. If $a^p=e$ then we know that $a\neq e$ and considering the cyclic subgroup of $G$, namely $H=\left \langle a\right \rangle$ $\Rightarrow$ $1<|H|\leqslant p<pm$

So we get two contradiction and it follows that $G$ is finite and its order is prime.

P.S. I think that my post is not duplicate because my solution somewhat is different from given duplicates. And this solution was created by me and it was important for me to understand is it correct or not.

• The Sylow theroems say that $G$ has Sylow-p subgroups for each prime factor of $|G|.$ – Doug M Nov 30 '17 at 17:27
• Good proof but your first sentence is wrong. it should say something like--Let $G$ be a group with no non-trivial subgroups, then we know $G$ must be finite by ..... Suppose that $|G| = n$ ... – Stephen Meskin Nov 30 '17 at 17:40
• This problem is a duplicate of at least two other problems: math.stackexchange.com/q/2219377/465208 and math.stackexchange.com/q/226365/465208 – Stephen Meskin Nov 30 '17 at 17:52
• @StephenMeskin, Maybe it is a duplicate but i would like to know is my solution correct or not? That's why i created this topic. – ZFR Nov 30 '17 at 17:58
• @StephenMeskin, let me ask you one question. The above statement states that if bla blah $\Rightarrow$ $G$-finite and it's order is prime. Hence the contradiction sounds: bla blah but $G$ infinite or it's order not prime. What's wrong? – ZFR Nov 30 '17 at 18:02

This is correct! Just a few comments, though:

If $G$ has no nontrivial subgroups $\Rightarrow$ $G$ is infiinite or $|G|=p$.

This is probably a typo, where you meant to say $|G| \neq p$?

There is also another small assumption, namely that the group $G$ itself is not trivial, for if it were trivial then it has no non-trivial subgroups, but its order is $1$, which is not prime.

EDIT: I went through the chat discussion linked in the comments under the question. The point Stephen is raising is that the first sentence of the proof is incorrectly worded. I chose to overlook this initially because I assumed English is not the OP's first language.

The point is that the first sentence:

Suppose by contradiction: If $G$ has no nontrivial subgroups $\Rightarrow$ $G$ is infiinite or $|G|=p$

does not express what you want to express in the most precise manner. Another way to read this implication is: "If if $G$ has no nontrivial subgroups then $G$ is infinite or $|G| = p$". This is already incorrect English, but it is also an incorrect way to state that you wish to prove the proposition by contradiction.

Here is how I would rephrase the first sentence.

$G$ is either infinite, or $|G| \neq p$ for any prime $p$, or $|G| = p$ for some prime $p$. We will show that if either of the first two conditions hold and $G$ has no non-trivial subgroups, then we arrive at a contradiction. This will show that if $G$ has no non-trivial subgroups, then $|G| = p$ for some prime $p$.

I hope this helps clarify the chat discussion, somewhat.

• Did I miss something? The group $\mathbb A_n$ for $n>4$ has no nontrivial subgroup and is not of prime order. – byk7 Nov 30 '17 at 17:30
• @byk7 $\mathbb{A}_n$ does not have any non-trivial normal subgroups, for $n > 4$. However, it does have non-trivial subgroups. – user279515 Nov 30 '17 at 17:31
• @Brahadeesh, let me ask you one question. The above statement states that if bla blah $\Rightarrow$ $G$-finite and it's order is prime. Hence the contradiction sounds: bla blah but $G$ infinite or it's order not prime. What's wrong? – ZFR Nov 30 '17 at 18:02
• @RFZ Exactly, either order is infinite or order is not prime, that is, either $G$ infinite or $|G| \neq p$. Isn’t it? – user279515 Nov 30 '17 at 18:03