Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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$A_5$ ?

share|cite|improve this question
See also… – David Speyer Apr 11 '12 at 16:13
On which page is this information? thanks a lot – user162343 Mar 31 '15 at 1:11
up vote 2 down vote accepted

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 $A_n$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.