# Is the question phrased properly? and is my proof correct? (An infinite alternating group is simple)

I'm interested in the following exercise from Dummut & Foote's Abstract algebra text (p. 151)

Let $D$ be the subgroup of $S_\Omega$ consisting of permutations which move only a finite number of elements of $\Omega$ (described in Exercise 17 in Section 3) and let $A$ be the set of all elements $\sigma \in D$ such that $\sigma$ acts as an even permutation on the (finite) set of points it moves. Prove that $A$ is an infinite simple group. [Show that every pair of elements of $D$ lie in a finite simple subgroup of $D$.]

I should note that in the beginning of the exercises section it says:

Let $G$ be a group and let $\Omega$ be an infinite set.

I suspect that the authors mean that $\Omega$ is assumed to be countably infinite. This is in light of the appearance of countable chains in the preceding exercise:

Prove that if there exists a chain of subgroups $G_1 \leq G_2 \leq \dots \leq G$ such that $G=\cup_{i=1}^\infty G_i$ and each $G_i$ is simple then $G$ is simple.

Assuming that $\Omega=\mathbb{N}=\{1,2,\dots\}$ is countably infinite, I have an attempt at the proof:

For $n \geq 5$ let $A_n$ be the subgroup of "even" permutations as defined in the exercise, which move only points of the set $\{1,2,\dots, n\}$. Then each $A_n$ is simple, and since the union $\cup_{n=5}^\infty A_n=A$ we find that $A$ is simple by the preceding exercise.

Showing that $A$ is infinite is easy, as it contains the infinite collection $\{(1 \;2)(3 \;n)\}_{n=4}^\infty$ of even permutations.

Must we assume that $\Omega$ is countable for this exercise? (Before answering this you might want to take a look at the exercises on page 151 of the text).

Thank you, and Happy Holidays!

P.S.

You can find the exercises with attempted solution (not by me) in section 4.6 here.

• @quid Which one? – user1337 Dec 23 '14 at 21:56
• @quid "The preceding exercise" is quoted after the text in bold. – user1337 Dec 23 '14 at 21:59
• Sorry for being dense here. I did not make the connection. – quid Dec 23 '14 at 22:01
• Just a remark regarding the "preceding exercise": More generally, it holds that any direct limit of simple groups is again simple (or trivial), which can be used (with some (but not much) effort) to prove the claim even for uncountable $\Omega's$. – Pavel Čoupek Dec 23 '14 at 22:28

There is no need to assume that $\Omega$ is countable.
Let $N$ be a nontrivial normal subgroup of $A$, so there exists $g \in N \setminus \{1\}$. We want to prove that $N=G$, so let $h \in G$, and we will prove that $h \in N$.
Let $\Delta$ be the set of all points moved by either $g$ or by $h$. So $\Delta$ is finite. If $|\Delta| < 5$, then adjoin finitely many more points to $\Delta$ to get $\Delta| \ge 5$. Let $H$ be the subgroup of $G$ consisting of elements that fix all points in $\Omega \setminus \Delta$. So $H$ is isomorphic to the alternating group on $\Delta$, which is simple. Now $g,h \in H$ and $g \in N \cap H$, which is a normal subgroup of $H$, so by simplicity we have $N \cap H = H$ and hence $h \in N \cap H$, QED.