Krdytkyu

I think the answer is $mathbb{Z}_n$ where $n$ is all the factors of $26$. Is this right?

$26 = 2cdot 13$. Both 2 and 13 are the prime factors of $26$, and cannot be reduced further.

So we know that So $$mathbb Z_{26} equiv mathbb Z_2 times mathbb Z_{13}$$

We have Lagrange Theorem to thank for proving that:

"...for any finite group $G,$ the order (number of elements) of every subgroup $H$ of $G$ divides the order of $G.$ The converse of the theorem applies only to cyclic groups.

More to the point: You need to prove that these are, and are the only, subgroups of $mathbb Z_{26}$:

$mathbb Z_{26},; {0},; {0,13},$ and ${0, 2,4,6,8,10, 12, 14, 16, 18, 20, 22, 24}.$

Notice the orders of of the groups listed: in order, we have $26, 1, 2, 13 $, which all divide $26$ (with no remainder).

And, indeed, we have that a cyclic group of order $n$ is isomorphic to the group $mathbb Z_n$

Not quite: Every subgroup of $mathbb{Z}_{26}$ is isomorphic to $mathbb{Z}_n$ for $n|26$, but such a $mathbb{Z}_n$ is not contained in $mathbb{Z}_{26}$ in a strict sense. The list of subgroups is the following:

• trivial subgroups: $mathbb{Z}_{26}$, ${0}$


• non-trivial subgroups: ${0,2,4,6,8,10,12,14,16,18,20,22,24}$, ${0,13}$.

Can you prove by yourself that these are indeed all subgroups? Do you see why the first non-trivial subgroup is isomorphic to $mathbb{Z}_{13}$ and the second one is isomorphic to $mathbb{Z}_2$?