Algebra Solutions: Hungerford I.6

2)(a) S_n is generated by the n-1 transpositions: (12),(13),...,(1n)
Proof-
Let \alpha \in S_n. We can write \alpha = (i_1i_2)(i_3i_4) \cdots (i_{r-1}i_r), where 2 \leq i_r \leq n. Note that each (i_ki_{k+1}) = (1i_k)(1i_{k+1})(1i_k). Using this we can write \alpha = (1i_1)(1i_2)\cdots (1i_{r-1})(1i_r)(1i_{r-1} \in \langle (12),(13),...,(1n)\rangle. Thus any permutation in S_n can be written as a product of the given transpositions.

(b) S_n is generated by the n-1 transpositions (12),(23),..., ((n-1) n)
Proof-
Let \alpha \in S_n. From part a we can write \alpha =  (1i_1)(1i_2)\cdots (1i_r) where 2 \leq i_r \leq n. Now consider (1i_k). If i_k = 2 then leave (1i_k) alone. If i_k > 2 then write (1i_k) = (1i_k-1)(i_k-1i_k)(1i_k-1). Repeat this process on each transposition of the form (1i_k-j) in the composition of \alphafor each j \geq as long as i_k-j>2. Doing this gives \alpha written as a product of transpositions of the form (i_k-1i_k) so that we are done.

12) The center of the group D_n is {e} if n is odd and \mathbb{Z}_2 if n is even.
Proof-
First we need to know all the normal subgroups in D_n. If n is odd, the nontrivial subgroups are the group of rotations, and all the subgroups of the group of rotations. If n is even we have two more normal subgroups, each isomorphic to D_{\frac{n}{2}}.
Let H be a normal subgroup. We have two cases, either H consists only of powers of the element a, or there is some a^ib \in H. Suppose that H = \langle a^k \rangle for some k dividing n. Then:
a^i(a^k)^ra^{-i} = (a^k)^r \in \langle a^k \rangle
b(a^k)^rb = b^2(a^{-2kr})=((a^k)^r)^{-1} \in \langle a^k \rangle because of the relation ba^r = a^{-r}b. And finally,
a^ib(a^k)^ra^ib = a^i(a^{-kr-i)}=a^{-kr} \in \langle a^k \rangle. Thus the subgroup is normal.
Now instead suppose there exists some a^kb \in H. Then we have a^{k+2}b = aa^k(ab) = a(a^kb) a^{-1} \in H since H is normal. This gives that (a^{k+2}b)(a^kb) = a^{k+2} a^{-k} = a^2\in H. This tells us that a^2 is a generator of H.
Next, a^{k-2l}b = (a^2)^{-l}(a^kb) \in H \forall l \in \mathbb{Z}. If k= 2m is even let l = m \Rightarrow b\in H and if k = 2m+1 is odd let l = m \Rightarrow ab\in H. Therefore either b is a generator or ab is a generator. So that H = \langle a^2,b \rangle or H = \langle a^2,ab \rangle.
Now that we know all the normal subgroups we can find the center, because it must be a normal subgroup. Recall the center is the set of all elements which can commute with any other element.
First we check H = \langle a^2,b \rangle. We have ab = ba \Rightarrow a^{-1} = a \Rightarrow a^2 = e, but n is larger than 2 so this is a contradiction. Thus this is not the center.
Next we check H = \langle a^2,ab \rangle. We have that:
a^2(ab) = (ab)a^2
a^3b = a^{-1}b \Rightarrow a^3 = a^{-1} \Rightarrow a^4 = e. But for n = 4, (ab)b = b(ab) \Rightarrow a = a^{-1} \Rightarrow a^2 = e again a contradiction so this is not the center.
Lastly, assume that \langle a^k \rangle is the center. Then we have ba^k = a^kb = ba^{-k} \Rightarrow a^{2k} = e \Rightarrow 2k = n \Rightarrow k = \frac{n}{2}. Thus the center can only be \langle a^{\frac{n}{2}} \rangle which the same as \mathbb{Z}_2 but only when n is even. If n is odd then the center is just the identity.

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