If $|G| = n < 60$, and $n$ is composite, then $G$ is not a simple group. Why?

Asked Dec 01 '21 at 11:19

Active Dec 01 '21 at 11:22

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If $|G| = n < 60$, and $n$ is composite, then $G$ is not a simple group. I am not totally sure how to solve this.

So far, I have tried thinking of every possible theorem I can think of.

A simple group is one where the only normal subgroups are trivial. If $G$ is group of prime order, then it must be a simple group since its only factors are $1$ and the prime order.

In this case, factors of $60 = 2^{2} \times 3 \times5 $. Now, according to Sylow's first theorem, it must contain $p$-groups of order of $4, 3$ and $5$.

And according to Lagrange's theorem, its subgroups must have order $1, 2, 3, 4, 5, 6, 10, 12, 15$.

I am not really sure how to proceed from here to show that indeed, the group cannot be a simple group. Any help would be greatly appreciated?

edited Dec 01 '21 at 11:22

Shaun

47,747

asked Dec 01 '21 at 11:19

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$A_5$ has order $60$ and is simple. – markvs Dec 01 '21 at 11:22
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$n<60$ you mean – David A. Craven Dec 01 '21 at 11:22
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I believe this question does not have a general or simple answer for all groups of composite order $<60$. We will need some case distinction to handle it (maybe not for every composite order but some classes of orders I can say). – ArsenBerk Dec 01 '21 at 11:26
@DavidA.Craven Yes,sorry about that – Dec 01 '21 at 11:27
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The answer may not be "simple", but it is very popular here at this site. There are many posts and detailed references. Another duplicate is this one, or this one, and there are more. – Dietrich Burde Dec 01 '21 at 11:32

If $|G| = n < 60$, and $n$ is composite, then $G$ is not a simple group. Why?

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