Isomorphic Groups

• Nov 26th 2006, 05:48 PM
PvtBillPilgrim
Isomorphic Groups
Really confused. I know all the definitions but I can't prove these. Can you just help me figure out the one condition or property that shows the following groups are not isomorphic?

I have this:
Prove that the following groups of order 8 are not isomorphic:
Z2 x Z2 x Z2 x Z2 x Z4
Z8
D4
Q.

Thanks.
• Nov 26th 2006, 05:57 PM
ThePerfectHacker
Quote:

Originally Posted by PvtBillPilgrim
Z2 x Z2 x Z2 x Z2 x Z4

I think you wanna write,
$\displaystyle \mathbb{Z}_2\times \mathbb{Z}_2 \times \mathbb{Z}_2$ and the group $\displaystyle \mathbb{Z}_2\times \mathbb{Z}_4$.
Hint.
How many abelian groups exist up two isomorphism of order 8?
Answer. Using the fundamental theorem of finitely generated abelian group we know that the only possibilities are the factorizations of 8 with prime powers. In that case,
1)$\displaystyle 8=2\cdot 2\cdot 2$ which yields,
$\displaystyle \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$.
2)$\displaystyle 8=2^2\cdot 2$ which yields,
$\displaystyle \mathbb{Z}_4\times \mathbb{Z}_2$
3)$\displaystyle 8=2^3$ which yeilds,
$\displaystyle \mathbb{Z}_8$
All these 3 groups are non-isomorphic. This also answers you next question.

Quote:

D4
The diheral group on 4 vertices is non-abelian :eek: . Those are.
Quote:

Q.
The group $\displaystyle \mathbb{Q}$ has greater cardinality then the ones given namely $\displaystyle \aleph_0$ those are of finite cardinality.
• Nov 26th 2006, 06:02 PM
PvtBillPilgrim
Yeah I did mean to separate those two.