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Suppose that a group $G$ of order $15$ has only one subgroup of order $3$ and only one subgroup of order $5$, then I need to prove that $G$ is cyclic.

If I can show that $\exists a\in G$ such that $|a|=15$, then the result is proved.
By Lagrange's theorem, every subgroup of order $3$ and $5$ is cyclic. Hence, let $H=\{e,a,a^2\}$ and $K=\{e,b,b^2,b^3,b^4\}$ be the only subgroups of $G$ of orders $3$ and $5$ respectively, such that $|a|=3, |b|=5$ and that $b\notin \langle a \rangle$
Clearly, $H\cup K=\{e,a,a^2,b,b^2,b^3,b^4\}\subset G$. Suppose that $x\in G$ but $x\notin H\cup K$, then by Lagrange's theorem, $|x|=1,3,5,15$. Clearly, $|x|\ne 1$. If $|x|=3$, then $x\in H$ and similarly if $|x|=5$ then $x\in K$. So $|x|=15$. Hence proved.

I can't believe it so I am looking for an example of such a group. How will this group look like? I just can't get past the following thought:
$G=\{e,a,a^2, b,b^2,b^3,b^4, ab, ab^2,ab^3,ab^4,a^2b,a^2b^2,a^2b^3,a^2b^4\}\tag{1}$

Will the group $G$ look like as in $(1)$? If yes, then
$(A)$ how can it be cyclic.
$(B)$ How to ensure closure i.e., whether for example $(ab)^2\in G$ or not?

Thanks for your time.

Koro
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    Hint: both subgbroups of order $3$ and $5$ must be normal. Verify that $HK$ is a subgroup. Then verify that $ab=ba$. – Arturo Magidin Jun 21 '20 at 22:01
  • @ArturoMagidin is this called the center of G? – Omar Shaaban Jun 21 '20 at 22:03
  • @ArturoMagidin, thanks a lot for your response sir. I understand that since groups of order 3 and 5 are cyclic (and hence Abelian) so they both are normal and hence $HK$ is a subgroup whence it follows that $ab=ba$. Can it be proven without using normal subgroups? – Koro Jun 21 '20 at 22:08
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    I did not mark as duplicate because it doesn't quite fit the criteria, but this is a very special case of a general fact: https://math.stackexchange.com/questions/5969/product-of-two-cyclic-groups-is-cyclic-iff-their-orders-are-co-prime – AnalysisStudent0414 Jun 21 '20 at 22:15

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Well, yes, you can certainly form the group $C_3 \times C_5$ which has the elements that you write above. The point is that... this group is cyclic! And it is generated by the product of the generators.

The reason is that $3$ and $5$ are coprime, and that is actually a necessary condition for this to happen (this = "the direct product of two cyclic groups being cyclic"): in fact, take $C_a = \langle x \rangle$ and $C_b = \langle y \rangle$, where $a$ and $b$ are coprime. Since the generators commute in $G=C_a \times C_b$, the order of $xy$ divides the least common multiple of the orders of $a$ and $b$ (since if $x^a=1$ then $x^{ak}=1$), so if $q$ is a multiple of both $a$ and $b$ then $x^q=y^q=1$ and hence $(xy)^q = 1$ (because, again, they commute).

But if the orders are coprime... the least common multiple is $ab$, the product of the orders! Since $G$ has order $ab$, then it follows that all elements of $G$ are powers of $ab$. In other words, $G$ is cyclic.

AnalysisStudent0414
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  • OK. So the idea is that generator for $G$ is product of generators of the two cyclic subgroups of order 3 and 5. – Koro Jun 22 '20 at 07:50