still the Michael Artin book, Chapter more on groups 2.6
Proof : To describe this isomprphism, we need to find a set $S$ of five elements on which $I$ operates. One such set consists of the five cubes which can be inscribed into a dodecahedron. (Omit the figure)
The group $I$ operates on this set of cubes $S$, and this operation defines a homomorphism $\phi: I \to S_5$, the associated permutation representation. The map $\phi$ is our isomorphism from I to its image $A_5$, to show that it is an icomorphism, we will use the fact that $I$ is a simple group, but we need very little infomation about the operation itself.
Since the kernel of $\phi$ is a normal subgroup of $I$ and since $I$ is a simple group. $\operatorname{ker}\phi$ is either $\{1\}$ or $I$. to say $\operatorname{ker}\phi=I$ would mean that the operation of $I$ on the set of five cubes was trivial operation, which it is not. There fore $\operatorname{ker}\phi=\{1\}$ and $\phi$ is injective, defining an isomorphism of $I$ onto its image in $S_5$.
I can understand that $I$ is simple group but how do we reach the conclusion that $\phi$ is injective ? I think I miss some knowledge I already learn before.
Let us denote the image in $S_5$ by $I$ too, we restrict the sign homomorphism $S_5\to\{\pm1\}$ to $I$, obtaining a homorphism $I\to\{\pm1\}$. If this homomorphism were surjective, its kernel would be a normal subgroup of I of order 30, this is impossible because $I$ is simple. Therefore the restriction is the trivial homomorphism, which just means that $I$ is contained in the kernel $A_5$ of the sign homomorphism. Since both groups have order $60$, $I=A_5$.
I didn't the last part, "Therefore the restriction is the trivial homomorphism, which just means that $I$ is contained in the kernel $A_5$ of the sign homomorphism"
what does the restriction mean ? is it related with the definiton of alternating groups? and how we come to the next step that "$I$ is contained in the kernel$A_5$ ?
Best Answer
The kernel of a group homomorphism $\phi:I\to S_5$ is a normal subgroup of $I$, but $I$ is simple, so $\operatorname{Ker}\phi$ must be either trivial, either $\{1\}$ or $I$. But it can't be all of $I$, since then the operation of $I$ would be trivial.
The restriction $f|_S$ of a any mapping $f:X\to Y$ to a subset $S\subset X$ is the mapping obtained by sending each $s\in S$ to $f(s)$ (note that $f_S$ is not defined on $X\setminus S$).
If we call the sign homomorphism $\sigma:S_n\to\{\pm1\}$, then it is pretty easy to see that $\operatorname{Ker}\sigma$ (which we know is a normal subgroup of $S_n$) has index $2$ in $S_n$, i.e., it is $A_n$.