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I have this following question from my class note on Sylow Theorem:

Show that a group of order 30 can not be simple.

For that I know the followings:
(1) A simple group is one that does not have non-trivial normal group,
(2) Group $G$ is p-group if there exists an integer $e$ such that $|G| = p^e,$
(3) A $p$-subgroup $H$ of $G$ is called Sylow p-subgroup of $G$ if $p \nmid |G/H|,$ and finally
(4) A subgroup $H$ of $G$ is a Sylow p-subgroup if there exist integers $e$ and $m$ such that $|G| = p^em,$ and $(e, m) = 1,$ and also $|H| = p^e,$

but unfortunately I don't know how to weave them into a solution, any help would therefore be very much appreciated. Thank you for your time and help.

PS. If you have choices of elegant versus dummy down-to-earth solutions, do please give me the latter. I know it would be tedious for you but you have a slowpoke turtle down here. Thank again for your patience.

POST SCRIPT : ~~~~~~~~~~~~~~~~~~~~~~~
I came across a very simple solution by John Beachy here, consisting of only 2 lines, which I adapted to fit group order of 30:

EDIT (j.p.) Unfortunately the linked solution cannot be easily adapted to this case here. I kept the text as far it was correct.

(1) Recall Sylow's Theorem: If $|G| = p^em; p, e, m \in \mathbb Z^+; (p, m) = 1;$ the number of $Syl_p (G) = s$, then $s \equiv 1\pmod p$ and $s \mid m.$
(2) For $|G| = 5 \cdot 2 \cdot 3$, suppose that $p = 5, e = 1$ and $m = 6.$ Therefore $s \equiv 1\pmod 5 $ and $s \mid 6,$ thus $s = \{1, 6 \}$. The 5-Sylow subgroup is normal if and only if $s=1$, and in this case $G$ is not simple.
(3) Now look at Nirdonkey's answer.

Hope this will be useful for some of you and thank you very much for those of you who have helped me.

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A.Magnus
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2 Answers2

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If $G$ is a group of order 30, then the number of Sylow 5-subgroups is congruent to 1 modulo 5 and divides 30, so it must be 1 or 6. And the number of Sylow 3-subgroups is congruent to 1 modulo 3 and divides 30, so it must be 1 or 10. If there is a unique subgroup of order 3 or 5, it is normal and we are done. If not, there are 6 subgroups of order 5 and 10 subgroups of order 3. But all of these groups must be cyclic, so every non-trivial element of such a subgroup is a generator of the subgroup. It follows that no non-trivial element of $G$ is contained in more than one of these subgroups. But then we have at least $6 \cdot 4 + 10 \cdot 2$ elements in $G$, which is a contradiction.

See also Group of order $|G|=pqr$, $p,q,r$ primes has a normal subgroup of order

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Nirvanacs
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  • I like your solution since my note covers the modulo $p$. One typo though: On the second line from above, "And the number of Sylow 5-subgroups is... " Don't you think it should be "... Sylow 3-subgroups... " instead? Thanks again and again. – A.Magnus Dec 22 '14 at 18:25
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By Cayley's theorem, a group $G$ of order $30$ is isomorphic to a subgroup of $S_{30}$ is such a way that no non-identity element of $G$ has any fixed point. There is an element $t$ of order $2$ in $G$ which is represented by a product of $15$ $2$-cycles, so as an odd permutation. The elements of $G$ which are represented as even permutations of $S_{30}$ form a normal subgroup of index $2$, so $G$ is not simple.

Geoff Robinson
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  • Why can you get an embedding into $S_n$ with no fixed points? – Nishant Dec 21 '14 at 23:05
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    Let $G$ act on $G$ by right translation. Only the identity fixes any point in this action. – Geoff Robinson Dec 21 '14 at 23:06
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    Hmm, so this proves that there is no simple group of order $2m$, where $m>1$ is odd. – Nishant Dec 21 '14 at 23:09
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    Yes, it does (this is a well known fact) In fact, it also shows that if a group $G$ whose order is of the form $2^{a}b$ (where $a$ is a positive integer, and $b$ is an odd integer) has cyclic Sylow $2$-subgroups, then $G$ is not simple ( for an element of order $2^{a}$ is represented by an odd permutation in Cayley's representation). – Geoff Robinson Dec 21 '14 at 23:12
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    @Nishant: I wanted to point out to you that Thompson's transfer lemma is another possible generalization of this line of argument, but then the first hit on google was this. You get the conclusion of your comment by taking $k=1$ in your question from July (looks a bit odd, but that's probably the reason you didn't make your observation back then). – j.p. Dec 22 '14 at 11:32
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    To be fair, I should point out that this was probably not the way the question was intended to be solved, since it doesn't really use Sylow. – Geoff Robinson Dec 22 '14 at 12:19