Group of order 15 is abelian

Question

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?

Answer 1

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)

Answer 2

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.

Answer 3

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...

Answer 4

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 $\gcd(n,\varphi(n))=1$. Here $\varphi$ is the Euler totient function. For $n=15$ this applies.

Answer 5

$G$ has exactly one normal subgroup of order $5$ and order $3$. Let us call them $N_3,N_5$ respectively. This follows from the Sylow theorems. As $|G| = 15$ but $|N_3| + |N_5| - |\langle 1_G\rangle| = 7$, there is a $g \in G$ that is not $1_G$, nor is it contained in $N_3$ or $N_5$. By closure of $G$ and Lagrange it follows that $\langle g \rangle$ = G.

As an exercise for the interested reader: If $|G| = p\cdot q$ for $p < q$ prime and $p$ not equal $1$ modulo $p$, then the above result holds. Even the counting argument holds, because $pq > p + q$ for $p,q$ prime and $p < q$.

Answer 6

(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).