# Elements of $\mathbb{Z}_6 / \mathbb{Z}_2$

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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• Since three is prime, there is only one group of order three up to isomorphism. Commented Aug 13, 2023 at 19:29

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.
• 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 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