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Why does order 135 imply nilpotent?

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@Vladhagen: Isn't "There are $5$ groups of order $135$" harder than the question asked? (I think so: it is known for all $n$ whether there is a non-nilpotent group of order $n$: there is a nice theorem here that takes maybe 2-3 pages to prove. How many groups there are of order $n$ is a very difficult question, on which many papers and several books have been written...) – Pete L. Clark Jan 19 '14 at 1:00
Touche. I was merely writing down some thoughts. But yes, finding out that all nonabelian groups of order 135 are isomorphic could be a complicated chore. – Vladhagen Jan 19 '14 at 1:14
@Vladhagen , BTW: $\;135=3^3\cdot 5\;$ so there are "only" $\;3\cdot 1=3\;$ abelian groups of order $\;135\;$ up to isomorphism. The other two are non-trivial semidirect products. – DonAntonio Jan 19 '14 at 4:53
Yes. So it is. Not on the top of my game today. – Vladhagen Jan 19 '14 at 5:15
@DonAntonio, the other two, that is, the non-abelian groups of order $135$, are direct products of a $3$-Sylow and a $5$-Sylow subgroup. The $5$-Sylow subgroup of order $5$ is cyclic, whereas for the (non-abelian) $3$-Sylow subgroup of order $3^{3}$ you have two possibilities, either exponent $3$ or exponent $9$. – Andreas Caranti Jan 19 '14 at 9:05

Let G be a group of order $135=3^{3}5$. Sylow theory says that if $p$ is a prime dividing the order of $G$, then, denoting the set of Sylow $p$-subgroups by $Syl_{p}(G)$, we have that $|Syl_{p}(G)| \equiv 1 \mod p$ and $|Syl_{p}(G)|$ divides $ |G|/|P|$, where $P \in Syl_{p}(G)$. $G$ is nilpotent if it is a direct product of its Sylow $p$-subgroups, which is true if and only if $|Syl_{p}(G)|=1$ for every prime $p$ dividing $|G|$. $$ $$ Let's calculate $|Syl_{3}(G)|$. We must have that $|Syl_{3}(G)| \equiv 1 \mod 3$ and $|Syl_{3}(G)| $ divides $ 5$ by the above. The only solution to this is $|Syl_{3}(G)|=1$. Now let's calculate $|Syl_{5}(G)|$. We must have that $|Syl_{5}(G)| \equiv 1 \mod 5$ and $|Syl_{3}(G)| $ divides $ 3^{3}=27$. So by the second statement, $|Syl_{5}(G)| \in \{1,3,9,27\}$. But only $1 \equiv 1 \mod 5$ out of these. So $|Syl_{5}(G)| = 1$ and hence $G$ is a direct product of its Sylow $p$-subgroups, so is nilpotent. $\square$

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Every group of order $135$ is nilpotent because $135$ is a nilpotent number. This note explains what that means. (It gives a simple arithmetic criterion on $n$ that is necessary and sufficient for every group of order $n$ to be nilpotent.)

Let me be upfront about the fact that the linked to paper, while short and nicely written, is not fully self-contained: at a key point it appeals to some results of Schmidt (that you could read about in a "serious" group theory text -- e.g. Robinson's GTM on group theory -- but not in the standard introductory algebra texts, so far as I know). So this answer is intended for "enrichment".

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