Questions in group theory

ehsan2004

I'm new in group theory and I want a little help.

1)Prove that the number of solutions of $x^3=e$ are odds.

2)Let $G$ be a group and $a,b \in G$ , $n \in \mathbb{N}$ and $xy^n=y^{n+1}x , yx^n=x^{n+1}y$ . Show that $x=y=e$ .

3)Let $G$ be a group and $A,B$ be two subgroups of $G$ and $|A|+|B|>|G|$ .Prove that $AB=G$ .

4)Prove that in any group $G$ the number of $a \in G$ which $a^2 \not= e$ are evens and show that if $G$ be a group ( $|G|=2k$ ) there exists an element $a \not=e$ s.t. $a^2=e$ .

**Posted:** Wed Mar 19, 2008 6:56 am

_________________
E.M.K

Soarer

1. Other than $e$ , for each $x$ such that $x^3 = e$ , $x^{-1}$ is another solution different from $x$ . Group $x$ with $x^{-1}$ , then you can see that the number of solutions is odd.

3. I suppose your G is finite. In this case, it means that if $|A| \ge |B|$ , then $2|A| > |G|$ , thus $|A| = |G|$ , and actually $A = G$ . Of course $AB = GB = G$ .

4. Similar to 1. If $a^2 \neq e$ , then $a \neq a^{-1}$ . Group $a, a^{-1}$ .
If $|G|$ is even. Then there are even number of elements $a$ with $a^2 = e$ (it's just the complement of the set defined just now). But $e$ is in this set, being even it has at least one other element, nontrivial, in this set. We are done.

**Posted:** Wed Mar 19, 2008 7:44 am

_________________
Hallo! Mein name ist Soarer und ich komme aus Hong Kong. Ich spreche Chinesisch, Englisch und ein bisschen Deutsch. Ich mag Mathematik!
Please correct my English or German.

vict85

2.
$xy = y(yx) = yx^2y = x^3y^2 = x^2(xy^2) = x^5y = x^4y^2x = x^3xy^2x = x^6yx = x^8y$
$xy = x^5y = x^8y\ \ \to \ \ x = x^5 = x^8$
$x^5x^{ - 1} = e \ \ \to \ \ x^4 = e$
$x^8x^{ - 5} = e \ \ \to \ \ x^3 = e$
$x^4 = xx^3 \ \ \to \ \ x = e$
$y^2e = ey \ \ \to \ \ y^2y^{ - 1} = e \ \ \to \ \ y = e$

**Posted:** Wed Mar 19, 2008 2:03 pm

LydianRain · Yang-Mills Theory Offline Joined: 02 Mar 2008 Posts: 785

Is it possible to use that solution for other values of $n$ ?

**Posted:** Wed Mar 19, 2008 2:26 pm

ehsan2004

**Posted:** Thu Mar 20, 2008 2:19 am

_________________
E.M.K

ZetaX

**Posted:** Thu Mar 20, 2008 9:25 am

_________________
The one and only place to get next year's IMO problems already now!

vict85

It wasn't the case n=1...
Anyway...

$xy^{n} = y^{n + 1}x$
$xy^{n}x^{ - 1} = y^{n + 1}$
$y^{n + 1} \in Orb(y^n)$

But $y^{n + 1} \in G_{y^n}$

$G_{y^n} \cap Orb(y^n) = \{ y^n \}$

So $y^{n + 1} = y^n\ \ \to \ \ y = e$

$x^{n + 1}e = ex \ \ \to \ \ x^{n + 1}x^{ - n} = e \ \ \to \ \ x = e$

**Posted:** Thu Mar 20, 2008 1:57 pm

LydianRain · Yang-Mills Theory Offline Joined: 02 Mar 2008 Posts: 785

What is $G_{y^n}$ ?

**Posted:** Thu Mar 20, 2008 5:26 pm

Soarer

**Posted:** Thu Mar 20, 2008 10:46 pm

_________________
Hallo! Mein name ist Soarer und ich komme aus Hong Kong. Ich spreche Chinesisch, Englisch und ein bisschen Deutsch. Ich mag Mathematik!
Please correct my English or German.

vict85

**Posted:** Fri Mar 21, 2008 5:22 am