# When can we “reduce” $(n\Bbb Z/m\Bbb Z)$?

$$(3\Bbb Z/6\Bbb Z) \cong (\Bbb Z/2\Bbb Z)$$ 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?

• 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 $6\mathbb{Z}$? – Arturo Magidin May 23 '19 at 5:41
• @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 $(2k)(2m)$. – Arturo Magidin May 23 '19 at 5:48
• $(2\mathbb{Z})/(6\mathbb{Z})$ 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. – Arturo Magidin May 23 '19 at 5:51

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.

• 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. – Sally G May 23 '19 at 11:50
• Totally agree. I should aviod non-unital rings. – eduard May 23 '19 at 12:33