Why is a group of order 135 nilpotent? Why does order 135 imply nilpotent?
 A: 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. $\square$
A: 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".
