I found this proof that no finite group is divisible.(here)
Let A be a finite divisible group then there are elements $x_k$ such that $x_k^k=1$ for each natural k.
Then that would mean there is an element of every order. Which is impossible since A is finite.
This answer doesn't convince me because we can let $e=x_k$ for each $k$. Is this reasoning correct, how would I go about proving this assertion?
Regards.
Best Answer
I agree with you. I think the linked proof is simply wrong, for exactly the reason you stated: there is indeed a solution of $x_k^k=1$ for every $k$, namely $x_k=1$. They say the words "Note that we may assume $x_k$ is minimal with respect to this property - that is, that $x_k$ has order $k$" in the linked proof, but they do not justify it, and I think it's total garbage/wishful thinking.
Put another way, the linked proof actually claims to show that no nontrivial finite abelian group is "divisible at the identity", but that's a false assertion.
Here's a proof that no nontrivial finite group (abelian or not) can be divisible: let $k$ be the maximal order of an element of the group ($k$ exists because the group is finite, and $k\ge2$ because the group is nontrivial), and let $a$ be an element of order $k$. I claim that the equation $x^k=a$ has no solution in the group.
Suppose, for the sake of contradiction, that there exists such an $x$. Let $j$ be the order of $x$, and note that $j\le k$ by the definition of $k$. If $j=k$, then $a = x^k = x^j = 1$, contradicting the fact that $a$ has order $k\ge2$. Otherwise $j<k$, and then $a^j = (x^k)^j = x^{jk} = (x^j)^k = 1^k = 1$, contradicting the fact that $a$ has order $k$.