$G$ is cyclic of order $10$, describe all quotients of $G$

Question

I have the following task:

$G$ is cyclic of order $10$, describe all quotients of $G$

My answer is: I know that $G \cong \mathbb{Z}/10\mathbb{Z}$ so we have 10 quotients

$[0]_{10}$ which is the neutral element

$[1]_{10}$ of order 10, $[2]_{10}$ of order 5, $[3]_{10}$ of order 10, $[4]_{10}$ of order 5, $[5]_{10}$ of order 4, $[6]_{10}$ of order 5, $[7]_{10}$ of order 10, $[8]_{10}$ of order 5, $[9]_{10}$ of order 10

Where [1],[3],[7],[9] are generators and cyclic and therefore abelian.

Is this correct and can more be said about these quotient groups?

Answer 1

No, it is not correct. If $G$ is generated by $a$, then its quotient groups are:

• $G/\{e\}\simeq G$;
• $G/\{e,a^5\}\simeq\Bbb Z_5$;
• $G/\{e,a^2,a^4,a^6,a^8\}\simeq\Bbb Z_2$;
• $G/G$, which is the trivial group.

Answer 2

Any quotient of a cyclic group is cyclic. One way to see it is that it is a homomorphic image (of the cyclic group).

Furthermore, any homomorphic image has order dividing the order of the group (first isomorphism theorem).

Finally it is easy to see that all the cyclic groups of orders dividing, in this case $10$, do occur as quotients.

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As far as what you did, a bunch of them are the same. So there are only four.