I can see that if $N$ is a group such that all $g \ne 1, g \in N$ have the same order, then this order is some prime $p$. Why is $N$ elementary abelian and of order $p^m$ for some $m$?
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12This is not true. For any prime $p \geq 3$ there is a non-abelian group of order $p^3$ where all non-identity elements have order $p$. – Tobias Kildetoft Jan 29 '13 at 15:31
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1Can we show just that $|N|=p^m$? – UbU Jan 29 '13 at 15:45
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2Yes, that follows for example from Cauchy's theorem. – Tobias Kildetoft Jan 29 '13 at 15:45
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1@user60108: Yes, because a group whose order is divisible by some prime number $q$ has an element of order $q$; therefore there cannot be any such prime factors $q\neq p$. – Marc van Leeuwen Jan 29 '13 at 15:47
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Elementary abelian? What does that mean please? – Julien Jan 29 '13 at 16:02
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This is true for $p = 2$. Let $a$ and $b$ two elements, then $(ab)^2 = (ba)^2 = e$ which means $abab = baba$.
Then, $ab(ab ab) = ba b(a a)b \Rightarrow ab = ba(bb) = ba$.
Damien L
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As Tobias Kildetoft mentioned in his comment, this need not be true. However, it is true if we assume that $N$ is abelian.
Let $N$ be a finite group such that
- $N$ is abelian
- all elements of $N$ have the same order.
Just from premise 2. we can conclude that $N$ has prime power order. For suppose $p$ and $q$ were two distinct primes such that $p\mid |N|$ and $q\mid |N|$. Then Cauchy's theorem implies that there are elements of order $p$ and $q$ contradicting 2. Thus there can be at most one prime dividing the order of $N$.
Assuming $N$ is abelian, we can conclude from the fundamental theorem of finite abelian groups (to use a sledgehammer) that $N$ is isomorphic to a direct product of cyclic groups with prime power order. If any of the factor groups were a cyclic group of order $p^i$ for $i$ greater than 1, we would know that there was an element of order $p^i$. This contradicts 2. as Cauchy's theorem still implies that there is an element of order $p$.