If G is a group of order n=35, then it is cyclic

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.

Thanks.

Solution 1:

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 $\pi$.

Solution 2:

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 $35 = 1 + |G_5| + |G_7|$

Note, you don't actually need Sylow to show that there are elements of order $5$ and $7$ here, because if $G$ is not cyclic, you have that $35=1+|G_5|+|G_7|$ and $|G_5|$ is a multiple of $4$ and $|G_7|$ is multiple of $6$, so it is not possible for either to be $0$.

Solution 3:

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.

Solution 4:

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

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

$|Syl_q(G)|\equiv 1 \pmod q$ and $|Syl_q(G)||p$ but $q\not \mid (p-1)$ 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 $C_p= \langle g \rangle$ and $C_q= \langle h \rangle$ then we have $o(gh)=lcm(o(g),o(h))=lcm(p,q)=pq$ and hence we have that $G$ is cyclic

Solution 5:

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.