How do I prove group of order 15 is abelian?

Is there any general strategy to prove that a group of particular order(composite order) is abelian?

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    $\begingroup$ 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. $\endgroup$ – Josué Tonelli-Cueto Sep 25 '11 at 11:44
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    $\begingroup$ 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. $\endgroup$ – Pete L. Clark Feb 16 '13 at 17:22

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.

  • $\begingroup$ 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 '13 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? $\endgroup$ – Lukas Juhrich Feb 3 '15 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. $\endgroup$ – YuiTo Cheng Aug 22 '19 at 12:00

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)

  • $\begingroup$ @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. $\endgroup$ – Pete L. Clark Sep 25 '11 at 20:47
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    $\begingroup$ 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. $\endgroup$ – Mikko Korhonen Dec 27 '11 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. $\endgroup$ – Pete L. Clark Dec 28 '11 at 0:25
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    $\begingroup$ 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. $\endgroup$ – Mikko Korhonen Dec 28 '11 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. $\endgroup$ – Jeppe Stig Nielsen Aug 12 '13 at 14:57

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...

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    $\begingroup$ Sorry, all. This just happened to be the first question in my candidacy exam (aka quals?) 24 years ago. I couldn't resist. $\endgroup$ – Jyrki Lahtonen Sep 25 '11 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 '18 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. $\endgroup$ – Jyrki Lahtonen Nov 10 '18 at 4:52

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.

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    $\begingroup$ 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 $\endgroup$ – Hans Parshall Sep 26 '11 at 21:43
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    $\begingroup$ The author even mentions that this result is "well known, but not widely known". $\endgroup$ – Hans Parshall Sep 26 '11 at 21:49
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    $\begingroup$ @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 $\gcd(n,\varphi(n)) = 1$. $\endgroup$ – Pete L. Clark Sep 26 '11 at 23:34
  • $\begingroup$ @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 tiny.cc/gp3s7 $\endgroup$ – Nicky Hekster Sep 27 '11 at 6:48
  • $\begingroup$ @Pete, thanks, yet another proof! $\endgroup$ – Nicky Hekster Sep 27 '11 at 6:49

Since $G$ contains an element $x$ of order $3$ and an element $y$ of order $5$. The map $f: \mathbb{Z_3} \times \mathbb{Z_5} \rightarrow G $defined by $f((i, j)) = x^i y^j$ is an isomorphism, provided $xy = yx$, and so $\mathbb{Z_3} \times \mathbb{Z_5} \cong \mathbb{Z_{15}}$

Hint : 1.o(G) = pq where p and q are primes with p<q 

And if p doesn't divide q-1, then G is cyclic. 2. Every cyclic group is an abelian group.


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