I need to prove that $(\mathbb{Z}_{2^n},+)$ has no faithful action on a set of size less than $2^n$. I tried to use division criterion since if an action is faithful (that is the kernel of the associate homomorphism is $0$) you have a injection of the initial group in $ S_{|X|}$, therefore you can use Lagrange theorem), but it did not work. I am relatively stuck so I will be thankful if a person can help me.
Thanks in advance!
Best Answer
Let $k < 2^n$, and say $\rho: \mathbb{Z}_{2^n} \to S_k$ is injective. Then, $\rho(1)$ has order $2^n$. Each cycle in the disjoint cycle decomposition of $\rho(1)$ must have order dividing $2^n$ (the order of $\rho(1)$) but not equal to $2^n$ (as $k < 2^n$). Then, the order of $\rho(1)$ must divide $2^{n-1}$, as each of the cycles in its disjoint cycle decomposition does. But this means $\text{ord}(\rho(1)) < 2^n$, which is a contradiction.