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