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- Jan 17, 2013

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This is Exercise 24 Chapter 3 Page 70

**Question**

*Suppose $n$ is an even positve integer and $H$ is a subgroup of $\mathbb{Z}_n$. Prove that either every member of $H$ is even or exactly half of the members of $H$ are even*

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**Attempt**

We consider some cases :

[1] The subgroup generated by the identity $0$ then $H=<0>=\{0\}$ hence all members are even .

[2] The subgroup generated by $1$ , $H=G$ and since the ordere of $G$ is even , we have half the members are even.

[3] The subgroup generated by $2$, $H=<2>=\{0,2,4,6 , \cdots , 2n\}$

hence all members are even.

[4] The subgroup generated by $n$ , $H=<n>=\{0,n\}$. If $n$ is even then subgroup contains even integers. If $n$ is odd half the members are even.

Since $\mathbb{Z}_n$ is cyclic every subgroup is cyclic. So it suffices to look at the subgroups generated by each even and odd element. I am taking this theorem from another latter chapter so I guess I shouldn't be able to use the properties of cyclic groups.

Any hint how to proceed ?