Order of $G$ is 35, are there elements of order 5 and 7 in G?

Question

I have a group of order 35 and I want to know if it contains elements of order 7 and 5. I know that it does and there is a proof that is much longer, but I wanted to know if the following worked to show that it does contain elements of order 5 and 7.

My Approach: Let $|G| = 35$. We know that any group of order 35 is cyclic so $G$ is isomorphic to $C_7$ x $C_5$ and $C_7$ contains an element of order 7 and $C_5$ contains an element of order 5. So, there are elements of order 5 and order 7 in $G$.

Correct, but the fact that $G$ must be cyclic is not trivial. — almagest, Apr 27 '16 at 15:23
This is correct, but the proof that any group of order $35$ is cyclic is rather more complicated than a direct proof that uses only, e.g. Lagrange's Theorem. — Travis Willse, Apr 27 '16 at 15:24
Thanks! It was proved in a book I have so I am allowed to use whatever is in the book. Could you also use the sylow theorems to show that there is one 5-sylow subgroup and one 7-sylow subgroup then these are cyclic and there must be an element of order 5 and element of order 7? — KhloeKard, Apr 27 '16 at 15:25
Yes, Sylow theorem also works. At the moment I cannot see a direct solution a la Travis! — almagest, Apr 27 '16 at 15:29

score 9 · Answer 1 · 2021-04-06T19:49:29.227

You can also show this using an elementary counting argument. First, note by Lagrange's theorem that every element must have order $1$, $5$, $7$, or $35$.

If there is an element of order $35$ then the group is cyclic, so there is some $g$ with order $35$. Then $g^5$ has order $7$, and $g^7$ has order $5$.

Now suppose there is no element of order $35$. Only the identity has order $1$, so every non-identity element must have order $5$ or order $7$.

If there is no element with order $7$, then every non-identity element has order $5$. Therefore $G$ is the union of $n$ subgroups of order $5$. Since $5$ is prime, each pair of subgroups intersects trivially, which means that we must have $|G| = 35 = 4n+1$, but there is no integer $n$ satisfying this equation.

Similarly, if there is no element with order $5$, then every non-identity element has order $7$. Then $G$ is the union of $n$ subgroups of order $7$. Since $7$ is prime, again the subgroups intersect trivially, so $|G| = 35 = 6n+1$. Again, there is no integer $n$ satisfying this equation.

We conclude that $G$ must contain at least one element of order $5$ and at least one element of order $7$.

I don't understand how you'd compare the order of $G$ with $4n + 1$, even if they intersect trivially, there'd be other elements like $xy$ where both have 5 as order. — Damn it My Foot, Mar 28 '21 at 15:03
@Karthik Yes, but in that paragraph we're assuming that every non-identity element of $G$ has order $5$, hence $xy$ would also have order $5$ unless $x$ and $y$ are inverses of each other in which case $xy$ would have order $1$. — , Mar 29 '21 at 05:22

score 2 · Accepted Answer · edited Apr 13 '17 at 12:19

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It follows by Cauchy's theorem, with $p=5$ and $p=7$. The proof of Cauchy's theorem is elementary. Furthermore we could use the result that all groups of order $pq$ with $p<q$ prime and $p\nmid q-1$ are cyclic - see here.

edited Apr 13 '17 at 12:19

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answered Apr 27 '16 at 15:30

Dietrich Burde

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Order of $G$ is 35, are there elements of order 5 and 7 in G?

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