Problem - Finite Abelian Group - Order of each element is = 2

• Sep 5th 2009, 11:49 PM
aman_cc
Problem - Finite Abelian Group - Order of each element is = 2
Problem: G = {a1,a2,a3,..,an} is a finite abelian group where order of each element (except of identity of-course) is 2.

Prove
1. Order of G is of the form $\displaystyle n = 2^k$ where k is a positive integer
2. a1.a2.a3....an = e (This is true for all values of n>2.)

My attempt for 1
If the group G, has a proper sub-group of order $\displaystyle 2^m$ then it has a sub-group (may not be proper) of order $\displaystyle 2^{m+1}$. This is how I did it. Let H be a proper sub-group of order $\displaystyle 2^m < n$. Thus, $\displaystyle \exists$ b $\displaystyle \in G$ but b not $\displaystyle \in H$. Consider $\displaystyle H_1= H \bigcup Hb$ where Hb is a coset of H. $\displaystyle H_1$ is a sub-group of order $\displaystyle 2^{m+1}$. Hence the result as there is a trivial sub-group of order 1 in G.

Any comments on this approach - is there a better way?
Any hints on 2?
• Sep 6th 2009, 01:59 AM
Taluivren
first note that you don't need to assume the group is abelian, because it follows from the fact that the order of each element (except identity) is 2.
Spoiler:

$\displaystyle abba=1$ implies $\displaystyle ba = (ab)^{-1} = ab$

then, assume G has more than one element and consider subgroup A1 of G generated by a nonzero element. Order of A1 is 2. But G/A1 is also a group in which every element is an involution... if it has more than one element you can take a quotient again...and so on. This process must terminate because G is finite and every step decreases number of elemts to one half.

for 2) we have n>2 so n >= 4. select two distinct nonzero elements u,v and consider subgroup H = {0,u,v,u+v} (these are 4 distinct elements).
then, partition G into cosets by H. You can rearrange the summation a1+...+an such that you first add elements from each coset a+H and then add those results all together. But summation of elements of coset a+H gives you (a+a+a+a)+ (0+u+v+(u+v)) = 0.
• Sep 6th 2009, 02:31 AM
aman_cc
Thanks a lot. Pretty neat solutions.
• Sep 6th 2009, 02:42 AM
Taluivren
You're welcome! Do you know the fundamental theorem of finitely generated abelian groups, or some less general version of it? It would solve these types of questions straightaway. But it is always fun to use elementary methods only. Your solution of 1) must be ok, too.
• Sep 6th 2009, 02:46 AM
aman_cc
No not really. New to the subject - just started a month back.
• Sep 6th 2009, 07:29 AM
ThePerfectHacker
If you have a bit more knowledge about groups you can fully classify what those groups are up to isomorphism. First, it must be abelian. Second, if $\displaystyle p$ is a prime number and divides the order of a group then there is an element of order $\displaystyle p$. Thus, we immediately see that $\displaystyle p\not | |G|$ for all odd primes and so this forces $\displaystyle |G| = 2^k$. Now by fundamental theorem of finite abelian group we get that: $\displaystyle G\simeq \mathbb{Z}_{2^{e_1}} \times ... \times \mathbb{Z}_{2^{e_r}}$. We must have $\displaystyle e_1=e_2=...=e_r=1$ because if $\displaystyle e_j > 1$ then you can find $\displaystyle X=(0,0,...,a,0,...0)$ (here $\displaystyle a$ is in the $\displaystyle j$ position) such that $\displaystyle X^2 \not = \text{identity}$. This would be a contradiction. Thus, $\displaystyle G$ must be isomorphic to a direct product of $\displaystyle \mathbb{Z}_2$ groups.