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Find the orders of $3 + \langle 6 \rangle$ and $2 + \langle 6 \rangle$ in $\mathbb{Z}_{15}/\langle 6 \rangle$. To what group is $\mathbb{Z}_{15}/\langle 6 \rangle$ isomorphic?

I know that:

$$\left|\frac{\mathbb{Z}_{15}}{\langle 6 \rangle}\right| = \frac{|\mathbb{Z}_{15}|}{|\langle 6 \rangle|} = \frac{15}{5} = 3$$

which is isomorphic to $\mathbb{Z}_3$.

How would I find the order of $3 + \langle 6 \rangle$ and $2 + \langle 6 \rangle$? And whichever has order of $3$ is the one that is isomorphic to $\mathbb{Z}_{15}/\langle 6 \rangle$?

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  • $\begingroup$ So both of their orders would be 3? And they are both isomorphic? $\endgroup$ – Kevin321 Apr 8 '14 at 4:16
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    $\begingroup$ Elements are not "isomorphic"... you should review the definitions. $\endgroup$ – user98602 Apr 8 '14 at 4:34
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The order of $g \in G$ is the smallest positive integer $n$ such that $g^n = e$.

As the group operation on $\mathbb{Z}_{15}$ is addition, we see that in $\mathbb{Z}_{15}/\langle 6\rangle$ we have $(3 + \langle 6\rangle)^n = 3n + \langle 6\rangle$. 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 $3n + \langle 6\rangle = \langle 6\rangle$. As $a + \langle 6\rangle = b + \langle 6\rangle$ if and only if $a - b \in \langle 6\rangle$, the order of $3 + \langle 6\rangle$ is the smallest positive integer $n$ such that $3n \in \langle 6\rangle$. Now note that $\langle 6\rangle = \{0, 6, 12, 3, 9\}$, so the smallest positive integer $n$ for which $3n \in \langle 6\rangle$ is $n = 1$. That is, $3 \in \langle 6\rangle$ so $3 + \langle 6\rangle = \langle 6\rangle$, i.e. $3+\langle 6\rangle$ is the identity element of $\mathbb{Z}_{15}/\langle 6\rangle$.

By a similar calculation, we find that the order of $2 + \langle 6\rangle$ is the smallest positive integer $n$ such that $2n \in \{0, 6, 12, 3, 9\}$. So $2 + \langle 6\rangle$ has order three.

Your calculation of the size of the quotient group $\mathbb{Z}_{15}/\langle 6\rangle$ is correct, but you should include some justification of why $|\langle 6\rangle| = 5$; for example $|\langle 6\rangle| = |\{0, 6, 12, 3, 9\}| = 5$. As there is only one group of order three, namely $\mathbb{Z}_3$, we have $\mathbb{Z}_{15}/\langle 6\rangle \cong \mathbb{Z}_3$.

An element of order three in $\mathbb{Z}_{15}/\langle 6\rangle$ is not isomorphic to $\mathbb{Z}_{15}/\langle 6\rangle$; we can say two groups are isomorphic, not an element of a group and a group. What you should have said is that an element of order three in $\mathbb{Z}_{15}/\langle 6\rangle$ is a generator of $\mathbb{Z}_{15}/\langle 6\rangle$. That is, the subgroup it generates is the whole group. For example, $2 + \langle 6\rangle$ is an element of $\mathbb{Z}_{15}/\langle 6\rangle$ of order three, so it generates the whole group; i.e. $\langle 2 + \langle 6\rangle \rangle = \mathbb{Z}_{15}/\langle 6\rangle$.

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I'd rather write $\bar{6}$ for the element considered in the group $\def\Z{\mathbb{Z}}\Z/15\Z$ and so on for the others.

The subgroup generated by $\bar{6}$ is the image of the homomorphism $\Z\to\Z/15\Z$ sending $1$ to $\bar{6}$, which is $(6\Z+15\Z)/15\Z=3\Z/15\Z$.

Therefore $(\Z/15\Z)/\langle\bar{6}\rangle$ is isomorphic to $\Z/3\Z$.

The isomorphism is the one sending $\bar{x}+\langle\bar{6}\rangle$ (where $x\in\Z$) to $x+3\Z$.

Thus $\bar{3}+\langle\bar{6}\rangle$ is sent to $0+3\Z$ (order $1$), while $\bar{2}+\langle\bar{6}\rangle$ is sent to $2+3\Z$ (order $3$).

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