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I would like to find an group isomorphism $f:\mathbb{Z}/17\mathbb{Z}\times \mathbb{Z}/561\mathbb{Z} \to \mathbb{Z}/51\mathbb{Z}\times \mathbb{Z}/187\mathbb{Z} $. By the fundamental theorem of a finite abelian group and the Chinese remainder theorem, we know that those groups are isomorphic, but I want to show it by constructing an isomorphism.

However, I don't know what the first step is. The only thing I know is that $f(0,0)=(0,0)$ since an isomorphism maps an identity element to an identity element.

Then I saw How to construct an isomorphism? and tried to mimic the way, like $f(x,y)=(x\mod{51},y\mod{187})$, but it is obviously not a surjection.

Now I am stuck here. Any help?

Kaira
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    I would try to insert an intermediate group, namely $\Bbb Z/17\times\Bbb Z/3\times\Bbb Z/187$. – Berci Dec 02 '20 at 06:45
  • See https://math.stackexchange.com/a/2226275/589 and https://math.stackexchange.com/questions/1724407/bbb-z-m-times-bbb-z-n-isomorphic-to-bbb-z-operatornamelcmm-n-times – lhf Dec 02 '20 at 10:07

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We introduce an intermediate group $\mathbb Z_{17}×\mathbb Z_3×\mathbb Z_{187}$. Represent an arbitrary element of this group as $(a,b,c)$ where the indices are residues modulo $17,3,187$ respectively.

There is an isomorphism from this group to the domain of $f$: $(a,b,c)\mapsto(a,187b+c)$. There is also an isomorphism to the codomain of $f$: $(a,b,c)\mapsto(3a+b,c)$. Put these two isomorphisms together and you have the required $f$.

Parcly Taxel
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