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I don't have enough reputation to comment on this neat answer so I ask here for some hints/answers. I take the set $A = \{0,1,2,3\}$ and so according to this answer, $A \sim \mathbb{Z}_4$. My question is: does that certain bijection allow to write the subset $\{0,2\}$ of $A$ is isomorphic to the subgroup $\{0,2\}$ of $\mathbb{Z}_4$?

I asked this because I don't know the explicit rule of the bijection. Thanks.

Hani
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  • Not necessarily because how do you know the identity is contained in ${0,2}$? – Peter Foreman Jul 19 '20 at 16:15
  • Please include all the information needed in a post. Do not ask the reader to have to follow a link to understand what is being asked. – fleablood Jul 19 '20 at 16:15
  • The answer in the question you link doesn't specify an explicit bijection. But if you take the most natural choice, that is, send $k$ to $k$ (mod 4), then yes ${0,2}$ is sent to a subgroup. – halrankard Jul 19 '20 at 16:23
  • halrankard: thanks for the hint. I see that. – Hani Jul 19 '20 at 16:32

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If $A\sim\Bbb Z_4$ means that $A$ is isomorphic to $\Bbb Z_4$, then the statement doesn't make sense, since $A$ is simply a set, not a group. Given any bijection $b\colon A\longrightarrow\Bbb Z_4$, you can use it to define a group structure on $A$ by transport of structure: if $a_1,a_2\in A$, $a_1\oplus a_2=b^{-1}\bigl(b(a_1)+b(a_2)\bigr)$. Then $(A,\oplus)$ is a group which is isomorphic to $(\Bbb Z_4,+)$. And $\{0,2\}$ may be or may not be a subgroup of $(A,\oplus)$; it depends upon the choice of $b$.

José Carlos Santos
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