# Non-abelian Group of order 8 problem

• Dec 3rd 2010, 06:49 AM
skyking
Non-abelian Group of order 8 problem
Heres the problem:

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

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

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

I was able to eliminate all options besides \$\displaystyle 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
• Dec 3rd 2010, 12:13 PM
tonio
Quote:

Originally Posted by skyking
Heres the problem:

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

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

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

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

\$\displaystyle 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

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