Let G be a cyclic group of order 24 then what is the total number of isomorphism ofG onto itself ??

Question

Let G be a cyclic group of order 24 then what is the total number of isomorphism ofG onto itself ??

Answer 1

Suppose $G = \{1,x,x^2, \ldots, x^{23}\}$ is the cyclic grounp and $h \colon G \to G$ is an isomorphism.

It is easy to show that $h(y)= 1 \iff y = 1$.

Now consider the the order of the element $h(x)$.

We have $h(x)^{24} = h(x^{24}) = h(1) = 1$.

So the order of $h(x)$ is no more than $24$. We claim the order is exactly $24$.

To prove that suppose $0<r<24$ is such that $h(x)^r = 1$.

Then we have $1= h(x)^r =h(x^r)$ and so $h(x^r) = 1$ which implies $x^r=1$ for some $0<r<24$, contradicting how $x$ has order exactly $24$.

We conclude that $h(x)$ has order exactly $24$.

Now ask yourself, which other elements of $G$ have order exactly $24$.

Answer 2

Let $C_n$ be the cyclic group of order $n$, generated by $x$. Then any endomorphism $f$ of $C_n$ is completely determined by where it sends $x$. If $f(x) = x^k$, then $f$ is surjective (and therefore injective) if and only if $k$ is coprime to $n$, hence the number of automorphisms of $C_n$ is $\varphi(n)$, the number of positive integers less than $n$ that are coprime to $n$.

Answer 3

Automorphism preserves order and here as the group is cyclic, any automorphism is completely dermined by finding where the generator has to be mapped. There are $\phi(24)=8$ elements of order 24 so there will be 8 total automorphism