Studying some concepts of group theory and after having read and trying to assimilate the concept of quotient group, I was wondering what structure would have the group $\mathbb{Z}_6 / \mathbb{Z}_2$. My reasoning was the following:
Using Lagrange's Theorem, we know that $|\mathbb{Z}_6 / \mathbb{Z}_2| = |\mathbb{Z}_6 : \mathbb{Z}_2| = \frac{|\mathbb{Z}_6|}{|\mathbb{Z}_2|} = \frac{6}{2} = 3$. Therefore, there are three equivalence classes, but taking into account that we are working with $\mathbb{Z}_2$, I think that all elements would be reduced modulo 2, and therefore I cannot see where the three equivalence classes appear. Am I using Lagrange's Theorem in an improper way? I would be grateful if someone could explain me.

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    – Shaun
    Commented Aug 13, 2023 at 19:27
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    $\begingroup$ Since three is prime, there is only one group of order three up to isomorphism. $\endgroup$
    – Shaun
    Commented Aug 13, 2023 at 19:29

2 Answers 2


You have to ask yourself: how do we view $\Bbb Z_2$ as a subgroup of $\Bbb Z_6$? The only possible way to do this is to identify $[0]$ with $[0]$ and to identify $[1]\in\Bbb Z_2$ with $[3]\in\Bbb Z_6$. Then for some $[n]\in\Bbb Z_6$, its image in $\Bbb Z_6/\Bbb Z_2$ corresponds to reducing $[n]$ modulo $[3]$. Explicitly, the cosets are $\{[0],[3]\},\{[1],[4]\},\{[2],[5]\}$; these are the three equivalence classes.

To be even more explicit, we are not really calculating $\Bbb Z_6/\Bbb Z_2$ in the sense that you're used to. Rather, if $\phi:\Bbb Z_2\to\Bbb Z_6$ is the homomorphism I described (taking $[1]$ to $[3]$) then we're calculating $\Bbb Z_6/\phi(\Bbb Z_2)$. But, because $\phi$ is an embedding, i.e. it corestricts to a group isomorphism $\Bbb Z_2\cong\phi(\Bbb Z_2)$, it would be extremely common to write $\Bbb Z_6/\Bbb Z_2$ since many things in mathematics are only considered "up to isomorphism" and we commit frequent abuses of notation in this vein.

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    $\begingroup$ Thanks for your answer, the visualization of $\mathbb{Z}_2$ as subgroup of $\mathbb{Z}_6$ was the point missing. Now I can see it much clearer $\endgroup$
    – Emmy N.
    Commented Aug 13, 2023 at 19:20

I have rethought this question and I have noticed that I could use First Isomorphism Theorem, finding a homomorphism $\phi : \mathbb{Z}_6 \rightarrow \mathbb{Z}_3$ with kernel $ker(\phi) = \mathbb{Z}_2$, induced by equivalence class


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