# Thread: Prove that group of order 15 is cyclic

1. ## Prove that group of order 15 is cyclic

Prove that a group of order 15 is abelian, and in fact cyclic.

Let $G$ be a group of order 15.
By Cauchy, $\exists a \in G$ such that period of $a = 3$.
By Cauchy, $\exists b \in G$ such that period of $b = 5$.

Now this is where I need help.

I think I want to say that: Since $G$ is a group, $ab \in G$. Then somehow show that the element $ab \in G$ has period 15 thus making it a cyclic group that generates $G$.

2. Originally Posted by Zalren
Prove that a group of order 15 is abelian, and in fact cyclic.

Let $G$ be a group of order 15.
By Cauchy, $\exists a \in G$ such that period of $a = 3$.
By Cauchy, $\exists b \in G$ such that period of $b = 5$.

Now this is where I need help.

I think I want to say that: Since $G$ is a group, $ab \in G$. Then somehow show that the element $ab \in G$ has period 15 thus making it a cyclic group that generates $G$.

Do you know the fundamental theorem of cyclic groups

Fundamental theorem of cyclic groups - Wikipedia, the free encyclopedia

3. Thanks! I will use that. I think I knew that but did not realize that it was a theorem.

4. along the lines of your original thought, if you can show G is abelian, then you can use ab to prove G is cyclic. in fact, it suffices to show that a and b commute.

now, what i would do is let G act on itself by conjugation: g(x) = gxg^-1. since the size of the conjugacy class of x is the index of the normalizer N(x) = {y in G: yxy^-1 = x},

if we have a non-identity element x of G fixed under conjugation, the size of its conjugacy class is 1, and so [G:N(x)] = 1, so N(x) = G. if N(x) = G,

then x is in the center of G, Z(G). but if G has a non-trivial center, it is of order 3,5 or 15. if G = Z(G), G is abelian.

clearly, |Z(G)| cannot be 3 or 5, because then we could take either a or b in the center, so a and b would commute, so ab would generate G, making Z(G) all of G.

the way ahead is proof by contradiction, suppose that G has no orbits of size 1, except the one containing the identity. the size of any orbit divides the order of G,

so the conjugacy classes must be of size 3 or 5. in fact, we must have 3 orbits of cardinality 3, and just one of cardinality 5.

consider c = aba^-1. if c is in <b>, then b = ebe^-1 = a^3ba^-3 = a^2(aba^-1)a^-2 = a^2(b^k)a^-2 = a(a(b^k)a^-1)a^-1 = a(b^(k^2))a^-1 = b^(k^3),

so k^3 = 1 (mod 5) --> k = 1, that is c = b, and thus a and b commute.

if c is not in <b>, then <c> is another subgroup of G with order 5. a similarly argument shows that either d = a^-1ba is in <b>, and thus a and b commute, or <d> ≠ <b>.

now if <c> = <d>, then c = d^m, so aba^-1 = a^-1(b^m)a, and multiplying on the left by a^2 and on the right by a^-2, b = a(b^m)a^-1 = c^m,

contradicting the fact that <c> ≠ <b>. this gives 12 distinct elements of order 5, which is too many to fit in the orbits (all elements in an orbit (conjugacy class)

have the same order).

so in all cases, a and b must commute, which means ab is of order 15.

@TheEmptySet: you can't use the fundamental theorem for cyclic groups unless you know that either a) G is abelian (this has to be proved) or b) that <a> and <b> are the ONLY subgroups of order 3 and 5. what i show above is that G can only have one subgroup of order 5, because if it has 2, it has 3, and that violates the class equation.

5. If you know Sylow's theorem(s), this is actually quite simple. Basically, use Sylow's theorem(s) to prove that there is only one Sylow 5-subgoup and only one Sylow 3-subgroup. This gives you that $G=C_5 \times C_3$. I will leave you to work out why...

Actually, this result generalises - any group of order pq where p and q are primes and q>p is a cyclic group if p does not divide q-1. Again, this is easily seen via Sylows theorem(s).

6. the Sylow theorems are far more powerful than order/conjugacy arguments, and usually make for nice pithy proofs.

short version: the number of subgroups of order 5 has to be of the form 1+5k, and also divide 15. 6 doesn't qualify, so that leaves just 1.

similarly, the number of subgroups of order 3 has to be of the form 1+3k and also divide 15, and 4 and 7 don't qualify.

i saw no indication, the original poster knew the Sylow theorems, often these types of questions are posed right after Cauchy's theorem and the class equation are introduced.

7. Originally Posted by Swlabr
If you know Sylow's theorem(s), this is actually quite simple. Basically, use Sylow's theorem(s) to prove that there is only one Sylow 5-subgoup and only one Sylow 3-subgroup. This gives you that $G=C_5 \times C_3$. I will leave you to work out why...

Actually, this result generalises - any group of order pq where p and q are primes and q>p is a cyclic group if p does not divide q-1. Again, this is easily seen via Sylows theorem(s).
In fact, it follows even easier from rep. theory haha

8. My professor skipped lecturing about Sylow theorems due to time. =(

9. Originally Posted by Zalren
My professor skipped lecturing about Sylow theorems due to time. =(

!!??!! It is completely understandable that he/she would skip the Sylow Theorems if the

course was about doodle drawing, the social life of ants in southern Australia in the late

18th century or the influence of Tolstoi in the life of hypos in the Nile River, but if the

course was about group theory (or abstract algebra) then it is utterly unforgivable not

to cover Sylow Thms, unless it was given as self study stuff.

If this isn't the case I'd recommend that teacher to take a swimm with hungry piranas

and giant octopuses...and no clothes allowed!

Tonio

Pd. If you're a math student it is very strongly recommended that you cover Sylow theorems

by yourself. It is very important and beautiful stuff.

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# prove that a group of order 15 is a abelian hence show that it is cyclic

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