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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

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  • $\begingroup$ And why don't wait @Ludolila to give you more details there and post this as a new question? $\endgroup$ – user26857 Mar 5 '13 at 21:04
  • $\begingroup$ You did this $1$ hour ago! Or do you pretend people answer you in a minute? (Ah, I see, this is a homework due at some moment.) $\endgroup$ – user26857 Mar 5 '13 at 21:09
  • $\begingroup$ 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? $\endgroup$ – Jim Mar 5 '13 at 21:54
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    $\begingroup$ People are crazy, @Jim. $\endgroup$ – Ragib Zaman Mar 5 '13 at 23:02
  • $\begingroup$ @jim: People use closure as a means of punishing questions that annoy them, whether or not the reason for closure is apt. $\endgroup$ – MJD Mar 6 '13 at 5:43
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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.

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  • $\begingroup$ +1 to Jim: Splitting morphism is a clean way (and equivalent to Andreas Caranti's upvoted argument). $\endgroup$ – Jyrki Lahtonen Mar 5 '13 at 21:43
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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.

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  • $\begingroup$ This is the same argument as the one given in the comments by @Ludolila here. $\endgroup$ – user26857 Mar 5 '13 at 21:07
  • $\begingroup$ 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. $\endgroup$ – Martin Brandenburg Mar 5 '13 at 21:08
  • $\begingroup$ @MartinBrandenburg, sorry, in my head I translate all groups to multiplicative form multiplicative. I will correct immediately, thanks. $\endgroup$ – Andreas Caranti Mar 5 '13 at 21:16
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    $\begingroup$ @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. $\endgroup$ – Andreas Caranti Mar 5 '13 at 21:21
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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 $G_1$.

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