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.
Best Answer
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 $33$.