Why isn't $D_3 \times \mathbb Z_{10} \cong D_{10} \times \mathbb Z_3?$

Question

I've looked at this through different angles and can't seem to come up with a good proof for this problem. Both groups have the same order $(60)$. I believe they also have elements of the same order $(1, 2, 5, 10, 15, 20)$.

Since $\mathbb Z_{10}\cong \mathbb Z_5 \times \mathbb Z_2$, by the fundamental theorem of finite abelian groups, I could turn the original problem turns into: $D_3 \times \mathbb Z_5 \times \mathbb Z_2 \ncong D_{10} \times \mathbb Z_3$, but I can't see how that would be helpful and every proof I've tried so far hasn't worked.

Any tips?

If you would search this site a bit, you could find ideas for a solution here. — Dietrich Burde, Jan 26 '20 at 13:37

score 6 · Answer 1 · answered Jan 25 '20 at 14:04

6

Consider the property

If $a$ has order $3$ and $b$ has order $2$, then $ab=ba$.

Only one of your groups has this property, the other doesn't. Hence they cannot be isomorphic.

answered Jan 25 '20 at 14:04

Hagen von Eitzen

374,180

Why isn't $D_3 \times \mathbb Z_{10} \cong D_{10} \times \mathbb Z_3?$

1 Answers1