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$(3\Bbb Z/6\Bbb Z) \cong (\Bbb Z/2\Bbb Z)$ can be easily shown by using the first isomorphism theorem, but I heard that we cannot say that $(2\Bbb Z/6\Bbb Z) \cong (\Bbb Z/3\Bbb Z)$.

Why not? And what are the conditions in which we can "simplify" the ring?

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  • $\begingroup$ You “heard” that? So the map that embeds $2\mathbb{Z}$ into $\mathbb{Z}$, followed by the projection to $\mathbb{Z}/3\mathbb{Z}$ is not additive, multiplicative, surjective, and has kernel $6\mathbb{Z}$? $\endgroup$ – Arturo Magidin May 23 at 5:41
  • $\begingroup$ @DionelJaime: Why not? $2k$ and $2m$ map to $2k+3\mathbb{Z}$ and $2m+3\mathbb{Z}$, respectively. Their product is $4km+3\mathbb{Z}$, which is the image of $(2k)(2m)$. $\endgroup$ – Arturo Magidin May 23 at 5:48
  • $\begingroup$ $(2\mathbb{Z})/(6\mathbb{Z})$ has three elements: $[0]$, $[2]$, and $[4]$. The map that sends $[4]$ to $1+3\mathbb{Z}$, $[2]$ to $2+3\mathbb{Z}$, and $[0]$ to $0+3\mathbb{Z}$ is additive. And $[4][4] = [4]$, $[4][2]=[2]$, $[2][2]=[4]$ is the multiplication in $2\mathbb{Z}/6\mathbb{Z}$, and the images satisfy the same. $\endgroup$ – Arturo Magidin May 23 at 5:51
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The confusion may rely on your definition of $\cong$.

If $\cong$ means group isomorphism then the formula $(2\mathbb Z/6\mathbb Z) \cong (\mathbb Z/3\mathbb Z)$ is true.

If the use of $\cong$ here means canonically isomorphic then the formula is false.

Recall that for a given class of algebraic structures (say rings, groups, fields, ...) two objects in the class are said to be canonically isomorphic if there is a unique isomorphism between them.

In your examples $3\mathbb Z/6\mathbb Z$ is a group of two elements and admits a unique isomorphism with $\mathbb Z/2\mathbb Z$, so they are canonically isomorphic. Indeed, in an isomorphism $$ 3\mathbb Z/6\mathbb Z\rightarrow \mathbb Z/2\mathbb Z $$ the class of $0$ needs to be mapped to the class of $0$. This leaves no freedom of choice for the image of $[3]$.

However, in the case $2\mathbb Z/6\mathbb Z\rightarrow \mathbb Z/3\mathbb Z$ there are $2$ isomorphisms (of groups) completely determined by the image of $2$. In this case any chocie $[2]\mapsto [1]$ or $[2]\mapsto [2]$ defines an isomorphism.

Further remarks Let $m,k$ be positive integers. Then $k \mathbb Z/mk\mathbb Z$ is a cyclic group of order $m$. Thus, it is isomorphic to $\mathbb Z/m\mathbb Z$ as groups. The number of possible isomorphisms $k \mathbb Z/mk\mathbb Z\rightarrow \mathbb Z/m\mathbb Z$ coincides with the number $\varphi(m)$ of generators of $\mathbb Z/m\mathbb Z$. Here $\varphi(m)$ denotes the Euler's totient function which is $\geq 2$ if $m \geq 3$.

Hence $k \mathbb Z/mk\mathbb Z\cong \mathbb Z/m\mathbb Z$ is always true as group isomorphism. However $k \mathbb Z/mk\mathbb Z$, $\mathbb Z/m\mathbb Z$ are only canonically isomorphic when $m=1$ or $2$.

EDIT:

Notice that the canonically part depends strongly on the class of structures you consider, as the comments suggest. The ring $2\mathbb Z/6\mathbb Z$ is unital (the unit is $4$) and $2\mathbb Z/6\mathbb Z$ is canonically isomorphic to $\mathbb Z/3\mathbb Z$ as unital rings.

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    $\begingroup$ Are we sure that in the non canonical case, [2] -> [1] is also an ismorphism? In @arturomargidin's comment, [2] was mapped to [2]. But if we map it to [1], I'm not sure if it's an isomorphism, because [2]^2=[4] so it is not idempotent. $\endgroup$ – SKYejin May 23 at 11:50
  • $\begingroup$ Totally agree. I should aviod non-unital rings. $\endgroup$ – eduard May 23 at 12:33

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