Prove that a group of order $77$ is cyclic.
I reached a step then got stuck.
My attempt:
Let $G$ be a group with $|G|= 77$. $G$ may have elements of orders $7$, $11$ and $77$ (divisors of $77$).
If $G$ has an element of order $77$, then we are done.
If $G$ has only elements of order $7$, then the number of elements of order $7$ is divisible by $\phi(7)= 6$, then we will have, due to the presence of the identity element of order $1$, $|G|= 77= 6k+1$, for some $k$. This yields that $76=6k$, but $6\nmid 76$.
Similarly if we suppose that $G$ has only elements of order $11$. We will work the same way until reaching $10\nmid 76$.
Hence we conclude that $G$ has elements of order $7$ and $11$.
Here I don't know how to continue.
I know that if $a,b \in G$, where $|a|= 7$ and $|b|= 11$ then $|ab|$ divides $lcm(7,11)= 77$, but how to show that there is an element of order $77$??
I should let you know that I didn't take the theorem saying:
$|HK|= \frac{|H||K|}{|H\cap K|}$, as I see it in similar proofs.
Best Answer
We're given that $|G| = 77 = 7 \cdot 11$. Since $7, 11$ are prime and $7 \not\mid (11 - 1)$, Sylow's Third Theorem implies that $G$ has exactly one subgroup of order $7$ and one of order $11$. So, $G$ contains exactly $7 - 1 = 6$ elements of order $7$, $11 - 1 = 10$ elements of order $11$, and $1$ element of order $1$ (the identity). Since $1 + 6 + 10 = 17 < 77$, $G$ must have elements with some order $\neq 1, 7, 11$, hence an element (in fact, $60$ elements) of order $77$.
The same argument applies mutatis mutandis for any group whose order is a product $p q$ of primes $p < q$ for which $p \not\mid (q - 1)$.