46
$\begingroup$

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?

$\endgroup$
3
  • 14
    $\begingroup$ Sylow's theorems for showing that the group is the direct product of its Sylow subgroups, then apply that every group of order $p$ or $p^2$ is abelian. That's the general strategy for small order groups. $\endgroup$ Sep 25, 2011 at 11:44
  • 8
    $\begingroup$ A belated +1 for the second sentence in your question. In the study of finite groups, I think such inquiries make the difference between busy work and real mathematics. $\endgroup$ Feb 16, 2013 at 17:22
  • $\begingroup$ Yes, there's a simple one based on the class equation, as soon as $p\nmid q-1$ (which is the case if $p=3$ and $q=5$). See here: math.stackexchange.com/a/4394327/1007416 $\endgroup$
    – user1007416
    Apr 11 at 19:40

4 Answers 4

45
$\begingroup$

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)

$\endgroup$
6
  • $\begingroup$ @Arturo: thanks, that is a very natural addition. Indeed, as I was typing up my answer I was wondering what could be said about the nilpotency class of a group which has order a $k$-power free nilpotent number. $\endgroup$ Sep 25, 2011 at 20:47
  • 1
    $\begingroup$ I believe (iii) is due to Tibor Szele (1947) (every group of order n is cyclic iff $\gcd(n, \phi(n)) = 1$). See T. Szele, Uber die endlichen ordnungszahlen zu denen nur eine Gruppe gehirt, Com- menj. Math. Helv., 20 (1947), 265-67. $\endgroup$ Dec 27, 2011 at 23:35
  • $\begingroup$ @m.k.: 1947 is unexpectedly late, but I might as well include this attribution until/unless something earlier turns up. Thanks. $\endgroup$ Dec 28, 2011 at 0:25
  • 3
    $\begingroup$ I agree that it is a bit surprising. But Szele is mentioned ("result by Szele") at for example oeis.org/A003277 and also projecteuclid.org/DPubS/Repository/1.0/… and all other references I have seen for this theorem. $\endgroup$ Dec 28, 2011 at 0:35
  • $\begingroup$ @m.k. Interesting. Here are all the relevant sequences on Sloane's where more info (references) can be found: nilpotent numbers A056867; abelian numbers A051532; cyclic numbers A003277. $\endgroup$ Aug 12, 2013 at 14:57
39
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Hi, I have one concern at the moment, say I have $A$, and $B$ are 2 normal subgroups of $G$, and $A \cap B = \{ e \}$. Will I always have that $AB \cong A \times B$? $\endgroup$
    – user49685
    Dec 9, 2013 at 17:28
  • $\begingroup$ As someone who has no background in group theory: What does $G=P_3P_5$ mean (better: how is it defined)? Is this some kind of combination of groups? $\endgroup$ Feb 3, 2015 at 0:04
  • $\begingroup$ Knowing $P_3$ is normal, it must be a union of conjugacy classes. Since the order of each conjugacy class is a divisor of 15, the only possibility is $3=1+1+1$, which implies $P_3$ is in the centre. $\endgroup$ Aug 22, 2019 at 12:00
17
$\begingroup$

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...

$\endgroup$
3
  • 5
    $\begingroup$ Sorry, all. This just happened to be the first question in my candidacy exam (aka quals?) 24 years ago. I couldn't resist. $\endgroup$ Sep 25, 2011 at 13:14
  • $\begingroup$ Sorry, not getting what does it imply after we have the statement - Therefore there are 15-5-3+1=8 elements that don't belong to a proper subgroup. Can you explain please? $\endgroup$
    – Taxicab
    Nov 9, 2018 at 12:08
  • $\begingroup$ @UnknownMathMan Let $x$ be one of those 8 elements not belonging to either proper subgroup. What is the cyclic subgroup it generates? How many elements in it? You can eliminate all but a single alternative. $\endgroup$ Nov 10, 2018 at 4:52
13
$\begingroup$

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.

$\endgroup$
5
  • 2
    $\begingroup$ This may be well-known but I did not know it. Thanks for the tip. I found a proof of this here - jstor.org/stable/2324062?seq=1 $\endgroup$ Sep 26, 2011 at 21:43
  • 1
    $\begingroup$ The author even mentions that this result is "well known, but not widely known". $\endgroup$ Sep 26, 2011 at 21:49
  • 2
    $\begingroup$ @Nicky: to say that there is only one group of order $n$ is equivalent to saying that every group of order $n$ is cyclic, so according to my answer this occurs iff $n$ is a squarefree "nilpotent number". Among square free numbers it is easy to see that the nilpotence condition holds iff $\gcd(n,\varphi(n)) = 1$. $\endgroup$ Sep 26, 2011 at 23:34
  • $\begingroup$ @Pete, thanks, yet another proof! $\endgroup$ Sep 27, 2011 at 6:49
  • $\begingroup$ @Hans, I found the result as a student in the late seventies and "published" this in the Problem section of the Nieuw Archief voor de Wiskunde, a Dutch math journal, see Problem NAvW 540, bit.ly/3dQmvBM $\endgroup$ May 18, 2020 at 14:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.