Showing that $U(2^n)$ is not cyclic for $n \ge 3$

Question

I'd appreciate a hint on how to solve the following problem:

Prove that $U(2^n) (n \ge 3)$ is not cyclic. ($U(m)$ is the group of positive integers $j \le m$ such that $\gcd(j,m)=1$, under multiplication $\mbox{mod}\,\,m$)

Since elements in $U(2^n)$ are coprime to $2^n$, $U(2^n)=\{1,3,5,...,2^n-3, 2^n-1\}$. I tried taking an arbitrary odd number, $1 \le 2k+1 \le 2^n$, and showing that $(2k+1)^{2^n} \not\equiv 1 \,\,\mbox{mod} \,\,2^n$, which would show that no element has order $2^n$ and therefore cannot generate $U(2^n)$ so that $U(2^n)$ is not cyclic.

I used the binomial theorem to expand $(2k+1)^{2^n}$ in general terms, and wanted to show that there is some coefficient that is not divisible by $2^n$, so that $(2k+1)^{2^n}\, \mbox{mod} \,\,2^n \not\equiv 1$. Problem is, the coefficients other than one are divisible by $2^n$, as far as I can see. I hope I'm not missing something obvious.

Outside of this, I'm afraid I'm out of ideas. How else can I go about this?

Thanks.

Answer 1

The first issue here is that the order of $U(2^n)$ is not $2^n$ - in fact, it is $\phi(2^n)=2^{n-1}$, where $\phi$ is Euler's totient function.

As a hint: note that the element $2^n-1$ is of order 2, since $$ (2^n-1)^2=2^{2n}-2^{n+1}+1\equiv1\pmod{2^n}. $$ Also, $$ (2^{n-1}+1)^2=2^{2n-2}+2^n+1\equiv1\pmod{2^n} $$ as long as $2n-2\geq n$, which holds since $n\geq 3$.

So, you have two (distinct) elements of order 2. Can that happen if $U(2^n)$ is cyclic?

Answer 2

By the way, it suffices to consider just $2^3$, since if its multiplicative group is non-cyclic, then necessarily the multiplicative groups mod $2^n$ for $n>3$ are non-cyclic, since they surject to the multiplicative group of $2^3$...

Answer 3

Here is a way of principle of mathematical induction to prove it. For n=3 it is simple to show that U(8) is not cyclic. Now assume that for any positive integer k>3, U(2^k) is not cyclic. We have to show that U(2^k+1) is not cyclic. The proof is by contradiction, we assume that U(2^k+1) is cyclic then it's every subgroup must be cyclic. Since, the integers which are prime to 2^k will also be prime to 2^k+1. Hence, U(2^k) will be a subgroup of U(2^k+1), but is not cyclic which is a contradiction to our assumption that U(2^k+1) is cyclic.

Answer 4

Suppose G=$U(2ⁿ)$ is cyclic. Then G must have atleast one element of order |U(2ⁿ)|=$\phi(2ⁿ)$=2

Now the number of elements of order d in a group is a multiple of $\phi(d)$

Hence the number of elements of order $2^{n-1}$ =k.$\phi$($2^{n-1}$)=k.$2^{n-2}$ Clearly $k.2^{n-2}$ $\leq$ $2^{n-1}$ (order of the group) Thus the possible values of k are 0,1&2.

• If $k=2$ then no. of such elements is $2.2^{n-2}=2^{n-1}$ (what do you think about this?) Each element of $U(2^n)$ has order $2^{n-1}.$ But identity 1 has order 1 !!. Hence $k≠2$
• If $k=1$ then no. of such elements is $2^{n-2}.$ i.e, exactly half the elements have order $2^{n-1}$ which is not possible.

$Hence \ k \ must \ be \ 0.$

That is, no element in $U(2^n)$ has order $2^{n-1}.$ Hence $U(2^n)$ is not cyclic.

Answer 5

Count elements of order two!

A cyclic group of even order has exactly one element of order $2$. For $n\geq 3$, both $-1=2^n-1 \bmod 2^n$ and $1+2^{n-1}\bmod 2^n$ have multiplicative order $2$, and they are distinct modulo $2^n$ since $1+2^{n-1}<2^n-1$ because $n\geq 3$ (they are equal when $n=2$).

In fact, for $n\geq 3$ there are always exactly three elements mod $2^n$ of order $2$: $-1, 2^{n-1}+1, 2^{n-1}-1 \bmod 2^n$. But this is not necessary in the argument above.