# To show from definitions , if $|G|=15$ then $G$ is isomorphic to $\mathbb Z_3 \times \mathbb Z_5$

How to show that any group of order $15$ is isomorphic to $\mathbb Z_3 \times \mathbb Z_5$ ? Please don't use results like "every group of order $15$ is abelian , cyclic " etc. just the definitions . Thanks in advance .

• What definitions do you have in mind? – Travis Sep 25 '14 at 15:39
• What are your thoughts on the problem ? it is not so elementary so that it can be solved straight from definitions - surely you can use theorems such as Lagrange theorem or even Cauchy – Belgi Sep 25 '14 at 15:39
• @Travis: Definition of $\mathbb Z_3 \mp \mathbb Z_5$ , first , second , third Isomorphism theorems , Cayley's and Extended Cayley's theorem , Lagrange's theorem ... etc . No conjugacy class or related facts – Souvik Dey Sep 25 '14 at 15:42
• What is Extended Cayley's Theorem? – Derek Holt Sep 25 '14 at 16:51
• @DerekHolt there seems to be an extended Cayley Hamilton theorem, but that is irrelevant in this context. – Marc Bogaerts Sep 25 '14 at 17:05

By Lagrange’s theorem, every element of $G$ (except $e$) has order $3,5$ or $15$. If there is an element of order $15$ we are done, so suppose there is none.
If there is no element of order $5$, every element of $G$ (except $e$) has order $3$. Let $x$ be one of them, and let $H=<x>$. By what you call "extended Cayley theorem", there is a homomorphism $f : G \to {\mathfrak S}_5$. Since every element of $G$ (except $e$) has order $3$, we see that every element in ${\sf Im}(f)$ (except $id$) has order $3$, so every element in ${\sf Im}(f)$ (except $id$) is a $3$-cycle. Clearly, all those $3$-cycles share the same support, because if we had two $3$-cycles $\alpha$ and $\beta$ with different supports, the product $\alpha\beta$ would have a support with size between $4$ and $5$, and so could not have order $3$. So the image of $f$ consists of the three powers of a certain $3$-cycle, and hence $|{\sf Im}(f)|=3$. We deduce $|{\sf Ker}(f)|=5$, contradicting the initial hypothesis that there is no element of order $5$.
All this shows that there is always an element of order $5$ (call it $k_1$). A similar and simpler argument shows that there is always an element of order $3$ (call it $h_1$). Let $H=<h_1>$ and $K=<k_1>$. So the orders of $H$ and $K$ are $3$ and $5$ respectively, which are coprime, so by Lagrange’s theorem $H\cap K=\lbrace e \rbrace$. It follows that the map $\phi : H\times K \to G, (h,k) \mapsto hk$ is injective, whence $G=HK$. By a similar argument, $G=KH$. So $G=HK=KH$, and there indices $a,c\in\lbrace 0,1,2 \rbrace$ and indices $b,d\in\lbrace 0,1,2,3,4 \rbrace$ such that $k_1h_1=h_1^{a}k_1^{b}$ and $k_1h_1^2=h_1^{c}k_1^{d}$. Note that none of $a,b,c,d$ can be zero because of $H\cap K=\lbrace e \rbrace$. If $a=c$, then $k_1H \subseteq h_1^{a}K$ and by induction we have $k_1^j H \subseteq h_1^{a}K$ for all $j\geq 1$, which becomes absurd when $j=5$. So $a\neq c$. By induction again, we have that $k_1^jh_1$ is in $h_1^{a}K$ when $j$ is odd, and in $h_1^{c}K$ when $j$ is even. Using this property for $j=5$, we see that $a=1$ and $c=2$. By induction on $i$, we then have $k_1^{i}h_1=h_1k_1^{bi}$ for any $i\geq 1$, and hence $k_1^{i}h_1^2=h_1^2k_1^{b^2i}, k_1^{i}h_1^3=h_1^3k_1^{b^3i}$. So we must have $b^3 \equiv 1 \ ({\sf mod}\ 5)$, which forces $b=1$. So $h_1$ and $k_1$ commute. This concludes the proof.
• From " all the 3-cycles have same support " how do you conclude that $|\Im (f)|=3$ ? Please elaborate – Souvik Dey Sep 29 '14 at 9:51
• @SouvikDey If this commun support is $S=\lbrace a,b,c \rbrace$ (say), there are exactly two $3$-cycles whose support is $S$ : $(a,b,c)$ and $(a,c,b)$. So ${\sf Im}(f)=\lbrace id,(a,b,c),(a,c,b)\rbrace$ – Ewan Delanoy Sep 29 '14 at 10:59
• Why does the existence of both an element of order $3$ and an element of order $5$ concludes the proof? You still need them to commute... – lhf Sep 29 '14 at 11:07
• @Ewan Delanoy : Why every element of $Im(f)$ (except $id$ ) is a $3$- cycle ? – user123733 Jan 10 '15 at 15:49