Here is a question from my Homework (Intro to College Algebra):

"Show that $\left( \mathbb Z_{2^n} \right)^*$ is not cyclic for any $n \ge 3$ (this is the multiplicative group consisting of the nonzero elements of $\mathbb Z_{2^n}$). [Hint: cyclic groups have unique subgroups of any order dividing the order of the group.]"

Note: $\mathbb Z_{2^n} = \{0, 1, 2, \ldots, 2^n-1\}$

PLEASE DON'T POST THE SOLUTION. I am quite frankly a bit confused with the problem itself. Instead, I would like clarification as to why this group would be cyclic for $n

Question

Here is a question from my Homework (Intro to College Algebra):

"Show that $\left( \mathbb Z_{2^n} \right)^*$ is not cyclic for any $n \ge 3$ (this is the multiplicative group consisting of the nonzero elements of $\mathbb Z_{2^n}$).  [Hint: cyclic groups have unique subgroups of any order dividing the order of the group.]"

Note: $\mathbb Z_{2^n} = \{0, 1, 2, \ldots, 2^n-1\}$

PLEASE DON'T POST THE SOLUTION.  I am quite frankly a bit confused with the problem itself.  Instead, I would like clarification as to why this group would be cyclic for $n < 3$... it doesn't make sense

anonymous · Answer

that i can answer i think
if $n=1$ you get $\mathbb{Z}_2$ which is clearly cyclic

anonymous · Answer

and if $n=2$ you get $\mathbb{Z}_4$ also cyclic

anonymous · Answer

actually now that i think about it, what exactly is $\mathbb{Z_n}^*$ ?

anonymous · Answer

oh nvm i see it is multiplicative  group of non zero elements of $\mathbb{Z}_n$
so $\mathbb{Z_2}^*=\{1\}$

anonymous · Answer

now i have a question
how is $\{1,2,3\} $ the non zero elements of $\mathbb{Z}_4$ a group under multiplication?  what is the inverse of 2?

anonymous · Answer

I have no idea haha...that's why the problem is confusing me... it seems blatantly obvious that $(\mathbb Z_{2^n})^*$ even for any $n$ wouldn't be cyclic anyway...