How Does Theorem 2.68 Explain Finitely Generated Groups in Algebra?

[Math Amateur]

I am reading Chapter 2: Commutative Rings in Joseph Rotman's book, Advanced Modern Algebra (Second Edition).

I am currently focussed on Theorem 2.68 [page 117] concerning finitely generated groups

I need help to the proof of this theorem.

Theorem 2.68 and its proof read as follows:View attachment 2698In the above text Rotman writes:

" ... ... if \(\displaystyle a \in A \) then \(\displaystyle \pi (a) \) is a word in the \(\displaystyle u\): there are \(\displaystyle e_j = \pm 1 \) with

\(\displaystyle \pi (a) = u_{i_1}^{e_1} ... \ ... u_{i_k}^{e_k}\) ... ... "

I am having real trouble following this proof ... can someone please explain the meaning of the above text and indicate why it follows.

Peter

[Thanks to Prove It for help with the Latex code in this post!]

 

[Deveno]

Here is another way to look at it:

We know the $u_i$ generate $A/S$. What are these elements? They are cosets:

$u_i = x_iS$, for some $x_i \in A$.

So an element of $A/S$ is a product of some of the $u_i$, and each of those is a product of some $x_i$ with a product of the $s$'s that generate $S$.

So we have $x_1S \subseteq \langle x_i,s_1,\dots,s_m\rangle$ <---$m+1$ generators "cover" this coset.

We only need $n$ of these cosets to generate all of $A/S$, so $\langle x_1,\dots x_n,s_1,\dots s_m\rangle$ contains every coset, that is, all of $A$.

Note: we might not need "all" of these, the generating set for $A$ COULD BE smaller, but we can at least find one this small, no matter what.

Let's look at an example, so this might be made a bit clearer.

Suppose $A = S_4$, and that the normal subgroup is $V = \{e,(1\ 2)(3\ 4),(1\ 3)(2\ 4),(1\ 4)(2\ 3)\}$, which is a normal subgroup (it is a union of conjugacy classes, the trivial conjugacy class of the identity, and the double transpositions. Recall that in $S_n$ conjugating preserves cycle type).

Now $V$ is the Klein 4-group, which is generated by any two non-identity elements, let's use $(1\ 2)(3\ 4)$ and $(1\ 3)(2\ 4)$.

Now $S_4/V \cong S_3 \cong D_3$ (the symmetry group of an equilateral triangle), and the latter group also has 2 generators. It can be shown (and I urge you to do so), that:

$(1\ 2\ 3)V$ and $(1\ 2)V$ generate $S_4/V$.

So the theorem tells us that $S_4 = \langle(1\ 2\ 3),(1\ 2),(1\ 2)(3\ 4),(1\ 3)(2\ 4)\rangle$. Is this true?

Well any permutation is a product of transpositions (the transpositions generate $S_n$, for any $n$), so it suffices to show that we can generate any transposition of $S_4$ with this set. It is immediate we can generate:

(1 2) and (3 4). (the latter transposition is simply (1 2)(1 2)(3 4)).

Now we also have (1 2 3)(1 2)(1 3 2) = (2 3) in the generated group, as well as:

(1 3 2)(1 2)(1 2 3) = (1 3). Since we have (1 3) in the generated group, we get:

(1 3)(1 3)(2 4) = (2 4) as well.

Finally, since all of $V$ is in the generated group, (1 4)(2 3) is generated by our set as well, since we have already established that (2 3) is generated by our set, it follows that:

(2 3)(1 4)(2 3) = (1 4) is in the generated group as well. So our set generates:

(1 2),(1 3),(1 4),(2 3),(2 4) and (3 4), which is all 6 transpositions of $S_4$.

Note that our generating set of 4 permutations is slightly more "efficient" than using the set of 6 transpositions.