Thread: 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.

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.

D4

The diheral group on 4 vertices is non-abelian . Those are.

Q.

The group $\displaystyle \mathbb{Q}$ has greater cardinality then the ones given namely $\displaystyle \aleph_0$ those are of finite cardinality.

Yeah I did mean to separate those two.