how to prove $G_1$ and $G_2$ aren't isomorphic? 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
 A: 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.
A: 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 $G_1$.
A: 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.
