7

Assume $$G_1=\mathbb Z_5 \times \mathbb Z_{5^2}\times \mathbb Z_{5^3}\times \mathbb Z_{5^4} \times\ldots$$ $$G_2= \mathbb Z_{5^2}\times \mathbb Z_{5^3}\times \mathbb Z_{5^4} \times \ldots$$ How do I prove $G_1$ and $G_2$ aren't isomorphic? I asked this question here Find distinct groups $G$ and $H$ such that each is isomorphic to a proper subgroup of the other and I received tree answer, but the answer of Ludolila will be complete when $G_1$ and $G_2$ are not isomorphic.

Thanks in advance

Community
  • 1
M.H
  • 11,498
  • 3
  • 30
  • 66
  • Just curious, there are a few votes to close with the reason being "too localized". But this site has hundreds of questions about how to do such and such integral and we don't close them. Why is this any different? – Jim Mar 05 '13 at 21:54
  • 2
    People are crazy, @Jim. – Ragib Zaman Mar 05 '13 at 23:02
  • @jim: People use closure as a means of punishing questions that annoy them, whether or not the reason for closure is apt. – MJD Mar 06 '13 at 05:43

3 Answers3

13

Note that the inclusion and projection maps $$\mathbb Z_5 \overset{\iota}{\to} G_1 \overset{\pi}{\to} \mathbb Z_5$$ compose to the identity on $\mathbb Z_5$. You want to show that such a pair of maps does not exist for $G_2$.

To show this assume we have $$\mathbb Z_5 \overset{\iota}{\to} G_2 \overset{\pi}{\to} \mathbb Z_5$$ and consider the element $x = \iota(1) \in G_2$. Show that it satisfies $5x = 0$ (because $\iota$ is a homomorphism). Then use this to show that there must be an element $y \in G_2$ such that $5y = x$. This then implies $\pi(x) = 0$ so $\pi\iota$ is not the identity.

Jim
  • 30,682
8

I may easily be wrong, but it seems to me that while all elements of order $5$ in $G_2$ are fifth multiples, not all elements of order $5$ in $G_1$ are.

PS I had written powers instead of multiples, because I had automatically translated in my head all involved factors as "a cyclic group of order $5$, a cyclic group of order $5^2$, etc.", and by default my groups are multiplicative. Sorry for the confusion, and thanks to @MartinBrandenburg for his comment.

Andreas Caranti
  • 68,873
  • This is the same argument as the one given in the comments by @Ludolila here. –  Mar 05 '13 at 21:07
  • Just for clarification (for me and perhaps for others): "Fifth power" refers to the group written multiplicatively. This is a bit confusing since we are dealing with abelian groups here. – Martin Brandenburg Mar 05 '13 at 21:08
  • @MartinBrandenburg, sorry, in my head I translate all groups to multiplicative form multiplicative. I will correct immediately, thanks. – Andreas Caranti Mar 05 '13 at 21:16
  • 1
    @YACP, and he was also thinking in terms of powers! Sorry, I gathered from the formulation of the present question that this was a matter left open in the other thread. – Andreas Caranti Mar 05 '13 at 21:21
4

obviously the reason they are not isomorphic is that $Z_5$ only occurs in one.

so how do we turn that into a proof? $Z_5$ is generated by an element of order 5, but we certainly have elements of order 5 in both groups so we will have to be able to say something different about the element of order $5$ in $Z_{5^2}$ too:

I think the key is that you can add one to it to get an element of order 24, i.e. there is an element $(1,1,1,1,1,1,1,\ldots) \in G_2$ that takes any element of order 5 to an element of order at least 25. Whereas there clearly is element with that property in $G_1$.