# For which $n$ is the group $\mathbb Z_n$ isomorphic to the subgroup of this group generated by $4$.

The exercise is as follows:

1. Consider the group $\mathbb Z_{24}$.
(a) For which $n$ is the group $\mathbb Z_n$ isomorphic to the subgroup of this group generated by the element $4$?

I'm a little confused by the wording. I understand that the element $4$ generates $$\{0,4,8,12,16, 20\}$$ but I am not sure how to relate this to the problem. If someone could show me how this relates to the question, or rephrase the question, it would be appreciated. Thanks.

• Hint: isomorphic groups have the same order, and the subgroup has order 5 as you can see. Feb 25 '16 at 3:19
• @NoahOlander Wait isn't the order $6$? Also, does that mean that the 'answer' is $\langle 4\rangle \simeq \mathbb Z_6$? Feb 25 '16 at 3:36
• Yeah, sorry, apparently I can't count! Feb 25 '16 at 4:03
• No prob, thanks @Noah. Feb 25 '16 at 4:17

The question is asking you to find one value of $n$ such that $\mathbb{Z}_n$ is isomorphic to the subgroup of $\mathbb{Z}_{24}$ generated by 4.

If all you need to do is find one such $n$, Noah's hint should suffice.

If you want to prove your choice of $n$ is correct, you will need to construct an isomorphism. Let $S$ be the subgroup generated by 4. You want to construct some $\phi : \mathbb{Z}_n \rightarrow S$ (you may switch the domain and codomain if you wish) such that:

• $\phi$ is a bijection
• For any $k_1$, $k_2 \in \mathbb{Z}_n$, we have $\phi(k_1 +_{n} k_2) = \phi(k_1) +_{24} \phi(k_2)$

where $+_{n}$ is addition modulo $n$. This will create an isomorphism between $\mathbb{Z}_n$ and $S$, thus showing they are isomorphic. Taking Noah's hint into account, you would just need to construct one such $\phi$.

• I don't think I have to construct it. I just need to state what $n$ is. And I believe $n = 6$. Is that correct? Feb 25 '16 at 3:42
• Noah says the order is 5, but I thought the order was the number of elements in set that is generated and the number of elements is 6. Am I mistaken? Feb 25 '16 at 3:51
• I believe it is worthwhile to convince yourself that $n = 6$ is correct, instead of just having me tell you whether it is or not. After all, if all you say is $n = 6$, you would need to back up that claim with some explanation. In this case, there is really no other option, so I will reveal that it is the correct answer, but understanding why is key to solving this problem. Edit: Yes, I believe Noah meant 6, not 5. Feb 25 '16 at 3:51
• I understand. I'm reviewing at the moment, but I'm pretty sure there are theorems\results that allow me to say $\langle 4\rangle \simeq S_6$. Further, I'm guessing there are theorems\results that say $S_6\simeq \mathbb Z_6$, so I can conclude that $\langle 4\rangle \simeq \mathbb Z_6$. I'll straighten it out soon enough. Hehe. Feb 25 '16 at 4:03
• Yep, I found the theorem. Thanks for the help. Feb 25 '16 at 4:17