User:SigmaJargon/Math5310

Prof. de Fernex Math 5310

Lemma A
If A is a subgroup of G and $$b\in G,\ b\notin A$$, then if $$b^n\in A,\ 0<n<o(b)$$ then $$(n,o(b))\ne1\,\!$$.

Proof: Say $$b^n\in A,\ 0<n<o(b),\ (n,o(b))=1\,\!$$. Then $$(b^n)=(b)\,\!$$, since $$o(b^n)=o(b)\,\!$$, and so $$b^n\,\!$$ is a generator of $$(b)\,\!$$. So since $$(b^n)=(b)\,\!$$, $$b\in A\,\!$$, which is a contradiction.

Lemma B
If A is a subgroup of G and $$b\in G,\ b^n\in A$$, then $$b^{(n,o(b))}\in A$$

Proof: Consider $$(b^n)\,\!$$. $$b^{x*n}\in (b^n)\,\!$$ for all x, and $$e=b^{y*o(b)}\in (b^n)\,\!$$ for all y. So $$b^{x*n}b^{y*o(b)}\in (b^n)\,\!$$. For some x, y, $$b^{x*n}b^{y*o(b)}=b^{x*n+y*o(b)}=b^{(n,o(b))}\,\!$$, so $$b^{(n,o(b))}\in (b^n)\,\!$$.

2.10.1
By Lemma A, since A is a subgroup of G and $$b\in G,\ b\notin A$$, if $$b^n\in A,\ 0<n<o(b)$$ then $$(n,o(b))\ne1\,\!$$. However, since $$o(b)=p\,\!$$, p prime, the only element that satisfies this condition is the identity, e. Thus, the only element in common between A and $$(b)\,\!$$ is e, and $$A\cap (b)=(e)\,\!$$

2.10.3.a
Say $$a^n\in N\,\!$$. We're guaranteed at least one such number, since $$a^0=e\in N\,\!$$. Then by Lemma B, $$a^{(n,o(a))}\in N\,\!$$, and so $$a^{(n,o(a))}N=N\,\!$$ by closure. Thus, $$o(aN)|(n, o(a))\,\!$$, and $$o(aN)|o(a)\,\!$$

2.10.3.b
Say $$o(aN)<o(a)\,\!$$. Then for some $$0<n<o(a)\,\!$$, $$a^nN=N\,\!$$. This implies $$a^n\in N\,\!$$, but this cannot be since $$(a)\cap N=(e)\,\!$$. So $$o(aN)\ge o(a)\,\!$$. Combined with $$o(aN)|o(a)\,\!$$, we conclude that $$o(aN)=o(a)\,\!$$

2.11.3
First, show $$C(x^{-1}ax)\subseteq x^{-1}C(a)x\,\!$$

If $$y\in C(x^{-1}ax)\,\!$$, show $$xyx^{-1}\in C(a)\,\!$$

Since $$y\in C(x^{-1}ax),\ y=x^{-1}axyx^{-1}ax\,\!$$, so $$xyx^{-1}=xx^{-1}axyx^{-1}axx^{-1}=axyx^{-1}a^{-1}\,\!$$

Thus, $$xyx^{-1}a=axyx^{-1}\,\!$$, and so $$xyx^{-1}\in C(a)\,\!$$, and $$C(x^{-1}ax)\subseteq x^{-1}C(a)x\,\!$$

Next, show $$x^{-1}C(a)x\subseteq C(x^{-1}ax)\,\!$$

If $$xzx^{-1}\in C(a)\,\!$$, show $$xzx^{-1}\in C(x^{-1}ax)\,\!$$

Since $$xzx^{-1}\in C(a),\ xzx^{-1}a=axzx^{-1}\,\!$$, or $$zx^{-1}ax=x^{-1}axz\,\!$$, and thus $$xzx^{-1}\in C(x^{-1}ax)\,\!$$. Finally, we have $$x^{-1}C(a)x\subseteq C(x^{-1}ax)\,\!$$, and so $$x^{-1}C(a)x=C(x^{-1}ax)\,\!$$

2.11.6
All p-Sylow groups are conjugate - that is, since P is a p-Sylow subgroup of G, if Q is a p-Sylow subgroup of G, then $$x^{-1}Px=Q\,\!$$. But since P is normal, $$x^{-1}Px=P\,\!$$, so P=Q. Thus, P is the only p-Sylow subgroup of G.

2.11.7
I hate to sound displeased with a problem, but we proved this last homework, problem 2.8.5. I'll attach the appropriate page for your viewing pleasure.

2.11.20
2.11.20) Consider $$p^n\,\!$$ divides $$|G|\,\!$$, such that n is as large as possible. Then $$|G|=p^nm\,\!$$, and since n maximal, p does not divide m.  Then by Sylow's Theorem, G has a subgroup P of order $$p^n\,\!$$.  If m=n, we're done.  If m<n, then we show $$p^m\,\!$$ is a subgroup of G by considering that since P is a group of order $$p^n\,\!$$, it contains a subgroup of order $$p^{n-1}\,\!$$.  Note that a subsubgroup is a subgroup, as it contains a subset of elements from the original group and meets the group axioms for the given operation.  If n-1=m, we're done.  If not, the process can be repeated as many times as needed, eventually showing that G contains a subgroup of order $$p^m\,\!$$