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Show that $\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/10\mathbb{Z}$ are not isomorphic

we have say that these two are not isomorphic means there does not exists bisection compatible with + and . i.e., $(f(x+y)=f(x)+f(y)$ and $f(x.y)=f(x).f(y)$

But how prove this

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The first group has exactly one element $(2,0)$ of order $2$, whereas the second group has two elements $(1,0), (0,5)$ of order $2$. If the groups were isomorphic, then they would have had same number of elements of possible orders.

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  • $\begingroup$ ....thank you! your genius $\endgroup$ – Inverse Problem Oct 10 '17 at 9:34
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The first group has an element of order $4$ but not the second.

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In the first one the equation $x^2-x=0$ has exactly $4$ solutions, while in the second one it has $8$.

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The two rings are not isomorphic because their additive groups are not isomorphic.

Indeed, the first group is cyclic but the second group is not, it has exponent $10$.

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The first one has nonzero nilpotent elements (e.g. $(2,0)$) but the second one has none.

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