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I've been asked to prove this.

In class we proved this when $n=15$, but our approached seemed unnecessarily complicated to me. We invoked Sylow's theorems, normalizers, etc. I've looked online and found other examples of this approach.

I wonder if it is actually unnecessary, or if there is something wrong with the following proof:

If $|G|=35=5\cdot7$ , then by Cauchy's theorem, there exist $x,y \in G$ such that $o(x)=5$, $o(y)=7$. The order of the product $xy$ is then $\text{lcm}(5,7)=35$. Since we've found an element of $G$ of order 35, we conclude that $G$ is cyclic.


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I can't see what's wrong with it, but I have a hard time believing that so many people would overlook this proof it it were correct. – user18297 Oct 26 '11 at 17:58
If $x$ and $y$ commute then the order of $xy$ is the lcm, but you can't assume this. – stopple Oct 26 '11 at 18:03
In fact, even if $x$ and $y$ have finite order, it need not be that $xy$ has finite order. – JavaMan Oct 26 '11 at 18:05
(following DJC's comment) e.g. We can express the Fibonacci matrix (with infinite order) as the product of two matrices with finite order (the orders are 2 and 3, if memory serves). – The Chaz 2.0 Oct 26 '11 at 18:07
So try applying this to a group of order 21. What goes wrong (since there is a non-abelian group of that order, which is therefore non-cyclic - as every cyclic group is abelian)? If you can't distinguish the cases, you don't have a proof. – Mark Bennet Oct 26 '11 at 18:07

As a concrete example, consider the single-cycle permutations $(1,2,3,4,5)$ and $(1,2,3,4,5,6,7)$, with orders $5$ and $7$, respectively. Their product is the cycle $(1,3,5,2,4,6,7)$ of order $7$.

On the other hand, pick two arbitrary axes in $\mathbb R^3$ and consider the groups of five-fold and seven-fold rotation symmetry about these axes. These are cyclic groups of orders $5$ and $7$, respectively, but the product of two elements, one from each group, is generally not a rotation through a rational multiple of $\pi$, and is thus generally of infinite order. You can see this by varying one of the axes; then the rotation angle of the product varies continuously with the orientation of the axis, and thus by the intermediate value theorem takes on irrational multiples of $\pi$.

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Here's a counting argument. If $G$ is not cyclic, then every element of $G$ is of order $1$, $5$, or $7$, and, as noted above, no element of order $5$ commutes with any element of order $7$.

Let $G_5$ be the elements of order $5$. We can see that $4\mid|G_5|$ by partioning $G_5$ into sets $\{g,g^2,g^3,g^4\}$.

On the other hand, given a $y$ of order $7$, we can partition $G_5$ into sets of $7$ elements, since we can separate $G_5$ into sets $\{g,ygy^{-1},y^2gy^{-2},...,y^6gy^{-6}\}$ (Note that, if $y^igy^{-i} = y^jgy^{-j}$, then $y^{i-j}g = gy^{i-j}$. So this must give $7$ distinct values, or some $y^{i-j}$ commutes with $g$, which would imply that $G$ is cyclic.)

So, $28\mid |G_5|$. Similarly, $30\mid |G_7|$. But $35 = 1 + |G_5| + |G_7|$

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Note, this works for any $|G|=pq$ where $p,q$ distinct primes and $(pq,(p-1)(q-1))=1$. On the other hand, if $p=2$, then there is a dihedral group of order $2q$ for all $q$. And if $p\mid (q-1)$, we can find a cyclic group of automorphism of $(\mathbb{Z}_q,+)$ of order $p$ by taking powers of an elements of $\mathbb{Z}_q^\times$ of order $p$, and take the semidirect product of the groups to get a non-cycle group of order $pq$. – Thomas Andrews Oct 26 '11 at 19:20

Here is another one,

Assume there are no elements of order $35$.

As $35 = 5 \cdot 7$, by Sylow's theorem there must be $1 \pmod 5$ subgroups of order $5$. And $1 \pmod 7$ of order $7$.

Now if there are $8$ subgroups of order $7$, then each one contains $6$ distinct elements $+$ id, so there would be $6 \cdot 8 = 48$ distinct elements in the group. This is clearly false, so there can only be $1$ subgroup of order $7$. So there are $6$ elements of order $7$.

Aside from the identity all the other elements must have order $5$. So there are $35 - 6 - 1 = 28$ elements of order $5$. We can split these into distinct subgroups each containing $4$ distinct elements $+$ id. So there are $7$ subgroups of order $5$.

But $7 = 2 \pmod 5$ so this contradicts Sylow's theorem. So there must be an element of order $35$ and $G$ is cyclic.

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Another explicit example:

Consider $$ A = \left( \begin{array}{cc} 1 & -1 \\ 0 & -1 \end{array} \right), \quad \text{and} \quad B = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right). $$ Then, $A^2 = B^2 = I$, but $$ AB = \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right) $$ has infinite order.

It should also be mentioned that if $x$ has order $n$ and $y$ has order $m$, and $x$ and $y$ commute: $xy = yx$, then the order of $xy$ divides $\text{lcm}(m,n)$, though the order of $xy$ is not $\text{lcm}(m,n)$ in general. For example, if an element $g \in G$ has order $n$, then $g^{-1}$ also has order $n$, but $g g^{-1}$ has order $1$. Joriki's example also provides a scenario where the order of $xy$ is not $\text{lcm}(m,n)$ in general.

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That's what I was talking about in my comment (though $AB$ here is the transpose...) – The Chaz 2.0 Oct 26 '11 at 18:54

I think we can prove this in more generality if we have that $|G|=pq$ with $p,q$ prime, $p> q$ and $q\not |(p-1)$

From the Sylow theorem we then have that the number if sylow-p groups is $1$ and then as:

$|Sylw_q|\equiv 1 mod q$ and $|Sylw||p$ but $q\not | p$ this then gives that the number of Sylow q-groups is also $1$

We then have that $G=C_p\times C_q$ and hence is abelian and so if we let $C_p=<g>$ and $C_q=<h>$ then we have $o(gh)=lcm(o(g),o(h))=lcm(p,q)=pq$ and hence we have that $G$ is cyclic

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As others have said, this proof does not work because you don't know <i>a priori</i> that it is abelian. It's entirely possible that $gh$ has infinite order (see: above matrix example). Let $n_p$ denote the number of Sylow $p$-subgroups of $G$. We know that $n_p \equiv 1 \pmod p$ and $n_p \mid 35$. We deduce from this that $n_5 = n_7 = 1$, so both the $5$-subgroup, $C_5$, and $7$-subgroup, $C_7$, are normal in $G$. By normality, we have that $C_5 C_7 < G$. Moreover, since $C_5 \cap C_7 = \{1\}$, $C_5 C_7 = C_5 \times C_7 \cong C_{35}$, and finiteness of $G$ implies that $G = C_{35}$. – JoeDub Aug 18 '14 at 23:57
@JoeDub I'm slightly confused. I have assumed that $|G|=pq$ and so all elements must be of finite order? – hmmmm Aug 19 '14 at 18:24

Without an appeal to the Sylow theorems, one can prove this through the Class Equation and the standard fact that if for a group $G/Z(G)$ is cyclic, then $G$ must be abelian.

Note that if $G$ is abelian, then by Cauchy's Theorem one can find an $x$, $y \in G$ of order $5$ and $7$ respectively. Since these elements commute, $xy$ has order 35 and $G$ must be cyclic.

So assume $G$ is not abelian. Then index$[G:Z(G)]$ cannot be $1, 5$ or $7$. Hence $Z(G)=\{1\}$. Now let us analyze the Class Equation: if $g \in G-\{1\}$, then $\{1\} \subsetneq\langle g \rangle \subseteq C_G(g) \subsetneq G$. Since $|G|=35$, it follows that $\langle g \rangle =C_G(g)$. Hence looking at conjugacy classes, $|Cl_G(g)|=7$ iff $o(g)=5$, and conversely $|Cl_G(g)|=5$ iff $o(g)=7$. A simple inspection shows that the only possible Class Equation is $$|G|=1+7+7+5+5+5+5,$$where the numbers denote the orders of the conjugacy classes. Now, since two subgroups of order $5$ either intersect trivially or are equal, the number of elements of order $5$ of $G$ must be a multiple of $4$. However, from the Class Equation it follows that there are $7+7=14$ elements of order $5$. A contradiction.

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