Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Yesterday, through working on a question Groups with only one element of order 2, Don antonio brought out a nice question within the comments:

The product of all the elements in an odd order group $G$ is always contained in the group's derived subgroup $G'$.

Honestly, I tried to link some facts for proving that but, they didn't work. Thanks for any hint for that.

share|cite|improve this question
up vote 8 down vote accepted

Let $\,G\,$ be a group, $\,G'=[G,G]=\,$ its derived or commutator subgroup, then we have the following:

(1) $\,G/G'\,$ is abelian, and thus

(2) Any product in $\,G/G'\,$ can be arranged at will by (1), and finally

(3) The product of all the elements in a group with an odd number of elements is in $\,G'\,$ since the product of their images in $\,G/G'\,$ is trivial.

share|cite|improve this answer
Thanks so much Don. (-: – Babak S. Nov 16 '12 at 19:55

By considering the quotient $G/G^\prime$, it suffices to show that the product of all the elements in an abelian group of odd order is trivial. This is true because for every $x$ that appears in the product, $x^{-1}$ also appears, and $x\ne x^{-1}$.

share|cite|improve this answer
Thanks Sean for the answer. – Babak S. Nov 16 '12 at 19:54

@Babak, there is a very neat but deep result here that goes a step further.

Let $G$ be a finite group of order $n$, say $G=\{g_1, g_2, ..., g_{n-1}, g_n\}$ and define $P(G)=\{g_{\sigma(1)} \cdot g_{\sigma(2)} \cdot \cdot \cdot g_{\sigma(n-1)}\cdot g_{\sigma(n)}: g_i \in G, i=1,... ,n$ and $\sigma \in S_n\}$, in other words $P(G)$ is the set of all possible products of $n$ different elements of $G$. (Of course the result of such product depends on the order of the elements, $G$ does not have to be abelian here!). Let $P$ be a Sylow $2$-subgroup of $G$.

If $P$ is non-cyclic or $\{1\}$ (that is $|G|$ is odd), then $P(G)=G'$.

If $P$ is cyclic, then $P(G)=xG'$, where $x$ is the unique element of order $2$ of $P$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.