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There is a part of a theorem that states "If d|n then there exists exactly one subgroup of order d. Then every element of order d in G also generates the subgroup."

Take Z4 < Z8 for example, Elements of order 4 in Z8 = {2, 6}. But Z4 = {0,1,2,3}, and <2> = {0,2,4,6} in Z8, which is clearly not Z4. What did I do wrong?

Unless <2> = {0,2,4,6} = 2Z8 is isomorphic to Z4 = {0,1,2,3}, but I don't know how to prove the isomorphism.

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  • $\begingroup$ Why don't you divide/multiply by $2$ remembering that the group operation is addition $\endgroup$ – Mark Bennet Feb 28 '17 at 18:23
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When you write $\;\Bbb Z_4\le\Bbb Z_8\;$ you're misleading, perhaps most of all yourself: the left one is integers modulo four whereas the right one is integers modulo eight: two completely different things!

What you probably mean is that there exists $\;H\le\Bbb Z_8\;$ such that $\;H\cong \Bbb Z_4\;$ . This much is true, for example (and following your example in some sense):

$$H:=\langle 2\rangle=\left\{0,\,2,\,4,\,6\right\}=\langle 6\rangle$$

Observe the above subgroup is taken from $\;\Bbb Z_8\;$ and thus the operation in it is addition modulo eight ...!

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