Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the product of $\mathbb{Z}_2$ and $\mathbb{Z}_5$ as subgroups of $\mathbb{Z}_6$?

Since $\mathbb{Z}_n$ is abelian, any subgroup should be normal. From my understanding of the subgroup product, this creates the following set: $\{ [0], [1], [2], [3], [4], [5] \}$ which has order 6 and is in fact $\mathbb{Z}_6$. However, the subgroup product formula $$ |\mathbb{Z}_2\mathbb{Z}_5| = \frac{|\mathbb{Z}_2||\mathbb{Z}_5|}{|\mathbb{Z}_2 \cap \mathbb{Z}_5|} = \frac{2 \cdot 5}{2} = 5 $$ I feel like I'm doing something wrong in the subgroup product, in particular understanding what closure rules to follow when considering the individual product of elements.

share|cite|improve this question
How is ${\bf Z}/5{\bf Z}$ a subgroup of ${\bf Z}/6{\bf Z}$? – anon Mar 18 '13 at 1:52
up vote 2 down vote accepted

You are possibly confused here since you write the elements of $\mathbb Z_n$ as $[0],[1],[2],\cdots $. This leads you to think that $\mathbb Z_5=\{[0],[1],[2],[3],[4]\}\subseteq \{[0],[1],[2],[3],[4],[5]\}= \mathbb Z_6$.This however is wrong. To understand why, recall that when you write $[1]\in \mathbb Z_5$ you mean that $[1]$ is the equivalence class of $1$ for the equivalence relation of $1$ modulo $5$. Similarly, when you write $[1]\in \mathbb Z_6$ you refer to the equivalence class of $1$ modulo $6$. These are very different sets and thus the element $[1]\in \mathbb Z_5$ is not in $\mathbb Z_6$. The same goes for all the other elements, so in fact $\mathbb Z_5\cap \mathbb Z_6=\emptyset$.

share|cite|improve this answer
Got it, thanks! – kordon Mar 18 '13 at 3:25
you're welcome. – Ittay Weiss Mar 18 '13 at 3:35

@Ittay Weiss, made you a complete illustration, but for noting a good point about the subgroups of $\mathbb Z$, we memorize:

If $m|n$ then $n\mathbb{Z}\leq m\mathbb{Z}$ (or $n\mathbb{Z}\lhd m\mathbb{Z}$).

share|cite|improve this answer
$\quad +^{{+^+}^+} \quad\ddot\smile\quad$ – amWhy Mar 18 '13 at 2:57

The only subgroups of $\Bbb Z_{6}$ are $\Bbb Z_{6}, \{0,2,4\}, \{0,3\}$, and $\{0\}$, corresponding to the divisors $1, 2, 3$, and $6$ of 6. There is no subgroup of order $5$. So there is no way that $\Bbb Z_{5}$ could be a subgroup of $\Bbb Z_{6}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.