# Isomorphic groups

• June 22nd 2009, 09:00 AM
Isomorphic groups
Trying to brush up on my group theory before next year...

One thing I've always had trouble with is proving two groups are (or in this case aren't) isomorphic.

So here's my Q.

Show that $C_8$ and $C_4 \times C_2$ are not isomorphic.

My thoughts... I'm thinking something along the lines of, an element of order 6 in the group $C_4 \times C_2$ is just the identity element since if you have $g \in C_4 \times C_2$ then $g^6 = g^{4+2} = g^4 \cdot g^2 = e \cdot e = e$? Whereas an element of order 6 in the group $C_8$ is not.

Think this is wrong so any help appreciated.
• June 22nd 2009, 09:08 AM
Swlabr
Quote:

Trying to brush up on my group theory before next year...

One thing I've always had trouble with is proving two groups are (or in this case aren't) isomorphic.

So here's my Q.

Show that $C_8$ and $C_4 \times C_2$ are not isomorphic.

My thoughts... I'm thinking something along the lines of, an element of order 6 in the group $C_4 \times C_2$ is just the identity element since if you have $g \in C_4 \times C_2$ then $g^6 = g^{4+2} = g^4 \cdot g^2 = e \cdot e = e$? Whereas an element of order 6 in the group $C_8$ is not.

Think this is wrong so any help appreciated.

There are no elements of order 6 in either group - let $G$ be an arbitrary group of order $n$, and let $g \in G$ be any element in $G$. Then $o(g) | |G|$ - that is, the order of any element of a finite group divides the order of the group.

Hint: prove that for $(g,h) \in C_n \times C_m$, then $o((g,h)) = max(o(g), o(h))$.
• June 22nd 2009, 03:47 PM
TheAbstractionist
Quote:

Originally Posted by Swlabr
Hint: prove that for $(g,h) \in C_n \times C_m$, then $o((g,h)) = max(o(g), o(h))$.

That is incorrect. You want the lcm, not the max.

@Deadstar: $C_8$ has an element of order 8. What is the maximum order of any element in $C_4\times C_2\,?$ Use Swlabr’s hint (corrected as above).
• June 22nd 2009, 10:08 PM
Swlabr
Quote:

Originally Posted by TheAbstractionist
That is incorrect. You want the lcm, not the max.

Thanks - I should be less hasty with my responses. (Doh)