Non-abelian Group of order 8 problem

**skyking** · December 3rd 2010, 06:49 AM

Heres the problem:

let $G$ be a non-abelian group of order 8

1. prove there exists an element of $G$ , call it $a$ such that it's order is 4.
2. Let $b$ be an element of $G$ such that $b$ is not formed by $a$ . Show that $a^{-1}=a^3=b^{-1}ab$ .
3. Show that if the order of $b$ is 2 then $G$ is isomorphic to $D_{4}$ , and if the order of $b$ is 4 then $G$ is isomorphic to the quatrernion group.

My solution so far:

I was able to show 1 easily by using legrange's theorem and eliminating the possiblities of 8,1,2 being the only order of elements.

Since $a$ is of order 4 it is obvious why $a^{-1}=a^3$ .

To show 2 I showed that the elements of $G$ must be $e,a,a^2,a^3,b,ab,a^2b,a^3b$ . Then my idea is to show by elimination that $ba^3=ab$ .

I was able to eliminate all options besides $ba^3=a^2b$ and this is where I'm stuck.

I'm not sure if this is the right approach at all or is there a much simpler and elegant way of showing this.

Thanks,

SK

**tonio** · December 3rd 2010, 12:13 PM

Originally Posted by skyking

Heres the problem:

let $G$ be a non-abelian group of order 8

1. prove there exists an element of $G$ , call it $a$ such that it's order is 4.
2. Let $b$ be an element of $G$ such that $b$ is not formed by $a$ . Show that $a^{-1}=a^3=b^{-1}ab$ .
3. Show that if the order of $b$ is 2 then $G$ is isomorphic to $D_{4}$ , and if the order of $b$ is 4 then $G$ is isomorphic to the quatrernion group.

My solution so far:

I was able to show 1 easily by using legrange's theorem and eliminating the possiblities of 8,1,2 being the only order of elements.

Since $a$ is of order 4 it is obvious why $a^{-1}=a^3$ .

To show 2 I showed that the elements of $G$ must be $e,a,a^2,a^3,b,ab,a^2b,a^3b$ . Then my idea is to show by elimination that $ba^3=ab$ .

I was able to eliminate all options besides $ba^3=a^2b$ and this is where I'm stuck.

$ba^3=a^2b\Longrightarrow a^3=b^{-1}a^2b\Longrightarrow a^2=(a^3)^2=b^{-1}a^2bb^{-1}a^2b=b^{-1}a^4b=1\Longrightarrow$ contradiction.

Tonio

I'm not sure if this is the right approach at all or is there a much simpler and elegant way of showing this.

Thanks,

SK

.

Thread: Non-abelian Group of order 8 problem

LinkBack

Thread Tools

Display

Non-abelian Group of order 8 problem

Similar Math Help Forum Discussions

The only non-abelian simple group of order 60 is A_5

For which n is there exactly one abelian group of order n?

Any group of order p^2 is abelian

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

Why is a group of order 3 abelian?

Search Tags