An abelian group $G$ of order $35$ with $g^{35}=e$ for all $g\in G$ is cyclic. I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.20.

Suppose that $G$ is an Abelian group of order $35$ and every element of $G$ satisfies the equation $x^{35}=e$. Prove that $G$ is cyclic. Does your argument work if $35$ is replaced with $33$?

There's definitely something I don't understand here. I am/was under the impression that, for any group $H$ and any $h\in H$, we have $h^{\lvert H\rvert}=e$; indeed: the cyclic subgroup $\langle h\rangle$ of $H$ has the same order as a group as the order $\lvert h\rvert$ of $h$ as an element of $H$; Lagrange's theorem then gives that $\lvert h\rvert$ divides $\lvert H\rvert$, so that then $h^{\lvert H\rvert}=e$.
So what gives?
I get the feeling that it's something obvious.
 A: My guess is that Gallian only proves Lagrange's theorem later.
If $(\forall g\in G):g^{35}=e$, then the order of every element of $G$ is $1$, $5$, $7$, or $35$. Of course, the goal is to prove that some element of $G$ has order $35$.
Suppose that every element other than $e$ has order $5$. Then every element of $G\setminus\{e\}$ belongs to some subgroup of order $5$. Each such subgroup will consist of $4$ elements of order $5$ plus $e$. But there's a problem here: $35$ is not of the form $4k+1$. So, some element from $G\setminus\{e\}$ must have order $7$ or $35$. And, by the same argument, not all elements from $G\setminus\{e\}$ have order $7$.
Therefore, some $a\in G$ has order $5$ or $35$ and some $b\in G\setminus\{e\}$ has order $7$ or $35$. If one of them has order $35$, we're done: $G$ is cyclic. Otherwise, $ab$ has order $35$ and, again, $G$ is cyclic.
And, clearly, this argument does not apply to $33$.
A: A more general version:
If $G$ is an abelian group of order $pq$, where $p<q$ are primes and $p-1$ doesn’t divide $q-1$, then G must be cyclic.
Proof:
The elements have possible orders $1,p,q,pq$.(by Lagrange theorem)
Now if there is an element of order $pq$, we’re done because we have found an element whose order is the order of the group hence the group is cyclic.
If there exists elements $a,b$: $|a|=p$ and $|b|=q$, then $|ab|$ divides $pq$. [Why? Because in an abelian group $$(ab)^{lcm(|a|,|b|)}=a^{lcm(|a|,|b|)}b^{lcm(|a|,|b|)}=e$$ so $|ab|$ divides $pq$]. But $|ab|≠p$ and $|ab|≠q $ so $|ab|=pq$. Again we're done.
If every non-trivial element has the same order say, $p$, then  we have $pq-1$ elements of order $p$, but $\phi(p)=p-1$ should divide $pq-1=p(q-1)+(p-1)$, so $p-1$ should divide $q-1$ which isn’t possible.
I have used the fact that: In a finite group, the number of elements of order $d$ is divisible by $\phi(d)$.
This doesn't work with 33 as $33=3\times 11$ but $(3-1)$ divides $(11-1)$
A: If $G$ is a group of order $n$, then $x^n=e$ by Euler's theorem for all $x\in G$. So this is an empty condition which is satisfied anyway. The point is that groups of order $pq$ are classified for primes $p<q$, depending on whether or not $p\mid q-1$. In one case all such groups are cyclic, in the other case there is a non-abelian group of that order.
Now check $(p,q)=(5,7),(3,11)$.
Reference: Let $p<q$ be distinct prime numbers and $G$ be a group with $|G|=pq$
