0

Let $(G, .)$ be a finite abelian group, $G=\{x_1, ..., x_n\}$ and let $x=x_1. \cdots. x_n$.

  1. Show that $x^2=e$
  2. Suppose $G$ has no element of order $2$ or that $G$ has more than one element of order $2$. Show that $x=e$.
  3. Suppose $G$ has a single element $y$ of order $2$. Show that $x=y$.

Progress: I have done part $1, 3$, and I shown that if $G$ has no element of order $2$ then $x=e$. For this part, It's easy since as $x^2=e$ and $o(x)\neq 2$ then $x=e$, but I'm stuck in the second part of part 2. Can someone help me?

mich95
  • 8,713
  • 2
    if $a_1,a_2$ are two distinct elements of order 2 then $a_1a_2$ is also a distinct element of order 2, so that $x=...a_1.a_2.(a_1a_2)...$ gives $e$. – r9m Mar 03 '14 at 02:23
  • But what if i have elements a1, a2 and a3,a4 or order 2? a1.a2 doesn't need to be different from a3.a4, I think –  Mar 03 '14 at 02:47
  • $a_1(a_1a_2)=a_2$ and $a_2(a_1a_2)=a_1$, so whenever we get 2 distinct elements of order 2, we get a triplet (a_1,a_2,a_1a_2) that cancels each other.If there are k elements of order 2, so that the product $a_1a_2..a_k$ has no cancellation of terms we get a $k+1$-tuple $(a_1,a_2,..,a_k,(a_1a_2..a_k))$ that becomes $e$ in $x$, and so on .. :) – r9m Mar 03 '14 at 02:56
  • I can't see why $a_1..a_k$ is different from each $a_i$ :( –  Mar 03 '14 at 03:07
  • sorry if I complicated it .. I meant if $a_i$'s are distinct from $a_1a_2...a_k$, (if not see that $a_1..a_{i-1}a_{i+1}..a_k=e$, so there is cancellation of terms in $a_1..a_k$), – r9m Mar 03 '14 at 03:17
  • that is really unnecessary though, in the set of all order 2 elements, say ${a_1,a_2,..,a_n}$, we pick two elements $a_1,a_2$, and argue that since $a_1,a_2,a_1a_2$ are all distinct elements of order 2, there is a $a_j$ ($j>2$) so that $a_j=a_1a_2$, that leaves us to deal with the set ${a_3,..,a_{j-1},a_{j+1},..a_n}$, we can continue this process to the point we have exhausted the set. – r9m Mar 03 '14 at 03:25

1 Answers1

1

Since $G$ is abeleian it is not important the order of multiplication.Let $A$ be set of elements of with order $2$ and $B=G-A$.Notice that since $B$ has no element with order $2$,$b\neq b^{-1}$ for all $b\in B$ except $e$ which is not important in multiplication.

Then,product of all elements in $G$ can be written as $$x=a_1a_2...a_k.b_1b_1^{-1}b_2b_2^{-1}...b_rb_r^{-1}$$

Thus,productof all elements of $G$ is equal to product of all elements of order $2$.

$1)$ Since $G$ is abelian $x^2=e$ as it is product of elemets of order $2$.

$2)$ if $A$ is emty,as above equation shows,$x=e$

$3)$ if $y$ is the only element with order $2$ then $x=y$

Now, let $H=<A>$ then observe that $H=A\cup \{e\}$ then from above construction multiplication of all elements of $G$ is equal to multiplication of all elements of $H$. Since $H$ is abelian and order of every nontrivial element is $2\implies H\cong Z_2xZ_2x..Z_2$

Assume $H$ has $m$ tupple since order of $A$ is more than one, $2\leq m$. Now think of sum of $(c_1,c_2,....,c_m)$ where $c_i\in \{0,1\}$ it is like sum of binary numbers.If you sum all possible forms you will get $(2^{m-1},2^{m-1},...,2^{m-1})$ and since $2\leq m$ ,it is zero in mod $(2)$ .

mesel
  • 14,862