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I'm trying to understand the previous thread (link) which finds the order of $3+\langle 6\rangle$ in $\mathbb{Z}_{15}/\langle 6\rangle$.

I was following the logic from that thread. First question is why in a group $(\mathbb{Z}_{15},+)$ we get:

$$(3+\langle6\rangle)^{n}=\underset{n\,times}{\underbrace{(3+\langle6\rangle)+(3+\langle6\rangle)+...+(3+\langle6\rangle)}}=3n+\langle6\rangle$$

An not $3n+n\langle 6\rangle$ (if we do get it and then why $n\langle 6\rangle =\langle 6\rangle$)?

Second question is about the line:

Now note that the identity of $\mathbb{Z}_{15}/\langle 6\rangle$ is $0 + \langle 6\rangle = \langle 6\rangle$, so the order of $3 + \langle 6\rangle$ is the smallest positive integer $3n + \langle 6\rangle = \langle 6\rangle$.

What is the identity of $\mathbb{Z}_{15}/\langle 6\rangle$? Is it $0$ or $\langle 6\rangle$? I feel like the author wanted to point out that $\langle 6\rangle$ is the identity element but why $0$ is not? Also later it said that $3+\langle 6\rangle$ is the identity element. How is it possible to have more then one identity element in a group?

Last question is why $\langle 6\rangle=\{0,3,6,9,12\}$? From this thread I learn that if order of a group is $15$ then total number of generators of group $G$ are equal to positive integers less than $15$ and co-prime to $15$. So for example $3$ and $9$ are not in $\langle 6\rangle$.

I feel confused and it feels like I miss the definition of the cyclic group (even if it in front of me). Will be glad for some clarifications. Also, sorry for wasting time on explaining basic things.

J. W. Tanner
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vesii
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  • $\langle 6\rangle$ is the smallest subgroup containing $6$. Keep adding $6+6+6+\cdots$, reduce modulo fifteen after each new term, and record the results. Those will be the elements of the subgroup. – Jyrki Lahtonen Feb 18 '19 at 15:15
  • We do have $\langle 6\rangle + \langle 6\rangle = \langle 6\rangle$, and not $\langle 6\rangle + \langle 6\rangle = 2\langle 6\rangle$. If you look at the definition of $+$ for subsets of an abelian group, and consider that $0\in \langle 6\rangle$, this fact will become clear. That being said, once you throw multiplication into the mix and get rings instead og groups, this would become a problem. Luckily, multiplication in a quotient ring is specifically defined to make this perceived problem not a problem. – Arthur Feb 18 '19 at 15:19

1 Answers1

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The addition in a quotient group $\mathbb{Z}_{15}/\langle 6\rangle$ is defined as $$(a+\langle 6\rangle) + (b + \langle 6\rangle) = (a+b) + \langle 6 \rangle$$ for $a,b \in \mathbb{Z}_{15}$. Hence no adding of $\langle 6\rangle$s.

On the other hand, equality in the quotient group is defined as $a + \langle 6\rangle = b + \langle 6\rangle$ if $a-b \in \langle 6\rangle$. The identity element of $\mathbb{Z}_{15}/\langle 6\rangle$ is $0 + \langle 6\rangle$, which is commonly denoted simply as $\langle 6\rangle$.

Let's determine $\langle 6\rangle$. Certainly $0,6 \in \langle 6\rangle$. Since a subgroup is closed under inverting we also have $-6 \in \langle 6\rangle$. But we have $-6 = 9$ in $\mathbb{Z}_{15}$ (the equality actually being congruence mod $15$) so $9 \in \langle 6\rangle$.

Since a subgroup is closed under addition, we have $12 = 6 + 6 \in \langle 6\rangle$. And then again $3 = -12 \in \langle 6\rangle$.

So far we got $\{0,3,6,9,12\} \subseteq \langle 6\rangle$. Now check that $\{0,3,6,9,12\}$ is indeed a subgroup of $\mathbb{Z}_{15}$, i.e. that the inverse of every element and the product of every two elements from the set stays in the set (working mod $15$ of course). Since $\langle 6\rangle$ is the smallest subgroup of $\mathbb{Z}_{15}$ containing $6$, we conclude $\langle 6\rangle \subseteq \{0,3,6,9,12\}$, and hence they are equal.

Finally, we see that $3 + \langle 6\rangle = 0 + \langle 6\rangle$ because $3-0 = 3 \in \langle 6\rangle$. Therefore $3 + \langle 6\rangle$ is indeed the identity element $\langle 6\rangle$ in $\mathbb{Z}_{15}/\langle 6\rangle$.

J. W. Tanner
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mechanodroid
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