Group of order 15 is abelian How do I prove that a group of order 15 is abelian?
Is there any general strategy to prove that a group of particular order (composite order) is abelian?
 A: Here is a 2000 paper of Pakianathan and Shankar which gives characterizations of the set of positive integers $n$ such that every group of order $n$ is (i) cyclic, (ii) abelian, or (iii) nilpotent.
Say that a positive integer $n > 1$ is a nilpotent number if $n = p_1^{a_1} \cdots p_r^{a_r}$ (here the $p_i$'s are distinct prime numbers) and for all $1 \leq i,j \leq r$ and $1 \leq k \leq a_i$, $p_i^k \not \equiv 1 \pmod{p_j}$.  Also, let us say that $1$ is a nilpotent number.
(So, for instance, any prime power is a nilpotent number.  A product of two distinct primes $pq$ is a nilpotent number unless $p \equiv 1 \pmod q$ or $q \equiv 1 \pmod p$.)
Then, for $n \in \mathbb{Z}^+$:
(i) (Pazderski, 1959) Every group of order $n$ is nilpotent iff $n$ is a nilpotent number.
(ii) (Dickson, 1905) Every group of order $n$ is abelian iff $n$ is a cubefree nilpotent number.
(iii) (Szele, 1947) Every group of order $n$ is cyclic iff $n$ is a squarefree nilpotent number.
For example, if $n = pq$ is a product of distinct primes, then $n$ is squarefree, so every group of order $n$ is nilpotent iff every group of order $n$ is abelian iff every group of order $n$ is cyclic iff $p \not \equiv 1 \pmod q$ and $q \not \equiv 1 \pmod p$.  In particular, every group of order $15$ is cyclic.
Addendum: This 2006 paper of T. Müller is a natural followup.  Rather than describing it myself, let me quote the MathSciNet review.

It is a popular problem to find for which positive integers n do all groups of order n have a given property (e.g., cyclicity, are abelian, etc.). The article under review contains a contribution to this problem which seems to have escaped notice.
Define a multiplicative function $\psi$ on the positive integers by letting $\psi(1)=1$, and $\psi(p^ν)=(p^{ν}−1)(p^{ν−1}−1)\cdots(p−1)$ if $p$ is a prime and $ν\geq 1$. The author proves that every group of order $n$ is nilpotent of class at most $c$ if and only if $\operatorname{gcd}(n,\psi(n))=1$ and $n$ is $(c+2)$-power free. Setting $c=\infty$ yields a result of G. Pazderski [Arch. Math. 10 (1959), 331--343; MR0114863 (22 #5681)] describing the case of nilpotency; and setting $c=1$ yields the classic result of L. E. Dickson [Trans. Amer. Math. Soc. 6 (1905), no. 2, 198--204; MR1500706] describing the case of abelianness. (Reviewed by Arturo Magidin)

A: Let $G$ be a group of order 15.  We know $G$ has subgroups of order 3 and order 5, say $P_3$ and $P_5$ from Sylow theory.  These must be cyclic (why?) so write $P_3 = \langle a \rangle$, $P_5 = \langle b \rangle$.
Using the lemma below, show $G = P_3P_5$.  Prove the lemma if it's not something you already know.
Lemma. For subgroups $H$ and $K$ of a finite group $G$, $|HK| = |H||K|/ |H \cap K|$, where $HK = \{hk \mid h \in H, k \in K\}$.
Using Sylow theory, show $P_3$ is normal.
Then $bab^{-1} \in \langle a \rangle$.  If $bab^{-1} = a$, we have $ba = ab$, so $G$ is abelian.  Observe $bab^{-1} \neq 1$ (why?).  The only "bad" possibility now is that $bab^{-1} = a^2$.
Suppose, to get a contradiction, that $bab^{-1} = a^2$.  Then $ba = a^2b$.  Using this identity repeatedly to fill in the $ \cdots $, show $a = b^5a = \cdots = a^2b^5 = a^2$.  But $a \neq a^2$, so this is a contradiction.
PS - Since $P_3$ and $P_5$ are both normal, you could instead argue that $G = P_3P_5$ implies $G \simeq P_3 \times P_5$.  In general, you can adapt this argument to show for primes $p,q$ with $p > q$ and $q \nmid p - 1$, every group of order $pq$ is abelian.
A: Hint: Any non-trivial subgroup is a Sylow subgroup. OTOH Sylow theorems tell that there is only one of order 3, and only one of order 5. Therefore there are 15-5-3+1=8 elements that don't belong to a proper subgroup, so...
A: In addition to the answers of Hans and Pete: it is well-known that if $n$ is a natural number, there is only one group of order $n$ if and only if $\gcd(n,\phi(n))=1$. Here $\phi$ is the Euler totient function. For $n=15$ this applies.
A: (Without: Sylow, Cauchy, semi-direct products, cyclic $G/Z(G)$ argument, $\gcd(n,\phi(n))=1$ argument. Just Lagrange and the class equation.)
Firstly, if $|G|=pq$, with $p,q$ distinct primes, say wlog $p>q$, then $G$ can't have $|Z(G)|=p,q$, because otherwise there's no way to accomodate the centralizers of the noncentral elements between the center and the whole group (recall that, for all such $x$, it must strictly hold $Z(G)<C_G(x)<G$).
Next, if in addition $q\nmid p-1$, then a very simple counting argument suffices to rule out the case $|Z(G)|=1$. In fact, if $Z(G)$ is trivial, then the class equation reads:
$$pq=1+kp+lq \tag 1$$
where $k$ and $l$ are the number of conjugacy classes of size $p$ and $q$, respectively.  Now,  there are exactly $lq$ elements of order $p$ (they are the ones in the conjugacy classes of size $q$). Since each subgroup of order $p$ comprises $p-1$ elements of order $p$, and two subgroups of order $p$ intersect trivially, then $lq=m(p-1)$ for some positive integer $m$ such that $q\mid m$ (because by assumption $q\nmid p-1$).
Therefore, $(1)$ yields:
$$pq=1+kp+m'q(p-1) \tag 2$$
for some positive integer $m'$; but then $q\mid 1+kp$, namely $1+kp=nq$ for some positive integer $n$, which plugged into $(2)$ yields:
$$p=n+m'(p-1) \tag 3$$
In order for $m'$ to be a positive integer, it must be $n=1$ (which in turn implies $m'=1$, but this is not essential here). So, $1+kp=q$: contradiction, because $p>q$.
So we are left with $|Z(G)|=pq$, namely $G$ abelian (and hence, incidentally, cyclic).
A: There's a unique normal Sylow subgroup of order $5$, because it's index is the smallest prime dividing the group's order.
It follows, after a couple details, that we have a semi-direct product $$G=\Bbb Z_5\rtimes \Bbb Z_3$$, which can only be trivial, because the orders of $\Bbb Z_5^×$ and $\Bbb Z_3$ are relatively prime.
