3

Let $G$ be a group s.t. every non-identity element has order 2. If $G$ is finite, prove $|G| = 2^n$ and $G \simeq C_2 \times C_2 \times\cdots\times C_2$

I know G is abelian since $ab = (ab)^{-1} = b^{-1} a^{-1} = ba$ for all non-trivial $a,b \in G$ so I have several questions remaining:

  1. How do I prove $|G| = 2^n$? I'd like to say we use induction to prove this but I'm at a loss as to where I would start.
  2. Why is $G \simeq C_2 \times C_2 \times\cdots\times C_2$?

I've read several answers to question similar to this yet unfortunately most of them involve Galois Fields and vector spaces, both concepts I'm unfamiliar with. I'd greatly appreciate an intuitive proof.

Michael Hardy
  • 1
Jonathan Lee
  • 51

2 Answers2

2

Suppose $G$ is finite, then we have $|G|= \prod_{i=1}p_{i}^{r_{i}}$. By the hypothesis and Cauchy's theorem, $p_{i} = 2$. So that $|G| = 2^{n}$.

Let $x, y \in G$, then $xy = 1$ or $(xy)^{2} = 1$ by the hypothesis and so we have $xy = yx$ since $g^{2} = 1 \implies g = g^{-1}$. Hence $G$ is an abelian group.

Write $G = \{g_{1}, g_{2}, ..., g_{n}\}$ where $|g_{i}| = d_{i}$. Let $H = \prod_{i=1}^{n}\mathbb{Z} \backslash d_{i}\mathbb{Z}$.

Then $\phi: H \rightarrow G: \phi (g_{1}, ..., g_{n}) = \prod_{i=1}^{n}g_{i}^{k_{i}}$ is a well-defined surjective map since $G$ is abelian but by order consideration, $\ker \phi = \{1 \}$. So we have the desired isomorphism since $d_{i} = 2$.

Daniel Akech Thiong
  • 4,129
  • I'm sorry but could you be more specific about how we can conclude that $|G| = 2^{n}$ by Lagrange's theorem and the hypothesis? – Jonathan Lee Dec 15 '15 at 04:39
  • Also, is there a way we can prove that G≃C2×C2…C2 without using the fundamental theorem for finitely generated abelian group? In other words, could we accomplish this proof using isomorphisms? – Jonathan Lee Dec 15 '15 at 04:42
  • I think in place of Lagrange's I should have said Cauchy's Theorem. If there is $p_{j}$ different from $2$, then $G$ has a non-identity element of order different from 2 which contradicts the hypothesis. So all the $p_{i}$ are just 2. Then the product is just $2^{\sum_{i=1}^{k}r_{i}} = 2^{n}$ – Daniel Akech Thiong Dec 15 '15 at 04:47
  • Okay, I understand that much. Now then, why do we get that $G \simeq C_{2} \times C_{2} \dots C_{2}$ as the primary decomposition of G and not anything else? We have not gone over the fundamental theorem for finitely generated abelian groups yet in class so I'm particularly curious about this one – Jonathan Lee Dec 15 '15 at 04:52
  • I have just written a completely new answer along the lines of what you have suggest: using isomorphism theorems and I have just posted it together with my earlier answer. – Daniel Akech Thiong Dec 15 '15 at 04:58
  • If anyone ever asks you to prove Cauchy's theorem for finite abelian groups, then this construction given in the answer is actually how one constructs an element of order prime p for every prime dividing the order of an abelian group. I thought I mention this since I have invoked Cauchy's theorem in the proof. – Daniel Akech Thiong Dec 15 '15 at 05:05
  • Can you please describe what the notations $p_i , r_i $ describe here? I can't realy see how you get the first conclusion. – Ron Abramovich Nov 12 '20 at 18:23
1

(1) $G$ is Abelian because $x y=x(x y)^2y=x x y x y y=x^2(y x)y^2=y x.$.. (2) For non-negative integer $n$, let $H_n$ denote a subgroup of $G$ with $|H_n|=2^n. $ It is easily seen that if $H_n\ne G$ and $x\in G\backslash H_n$ then $H_{n+1}=H_n\cup x H_n$ is a subgroup of $G$ with $|H_{n+1}|=2^{n+1}$....(3) Now $H_0=\{1\}$ exists. Take a sequence $x_0,...,x_n$ in $G$ and a sequence $H_0,...,H_n$, with $x_0=1$, $H_0=\{1\}$, such that for $0\leq i< n$ we have $x_{i+1}\not\in H_i$ and $H_{i+1}=H_i\cup x_{i+1}H_i,$ and where $n$ is as large as possible. By part (2) we must have $H_n=G$. This gives us $|G|=2^n$....(4) In the non-trivial case $n\geq 1$, observe that $H_n$ (which is $G$) is isomorphic to $\prod_{j=0}^{j=n-1}(H_{n-j}/H_{n-j-1})$ and each member of this product is a two-element group.

DanielWainfleet
  • 59,529