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 $(n,o(b))\neq 1\,\!$.

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 $(n,o(b))\neq 1\,\!$. 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 $0<n<o(a)\,\!$, $a^{n}N=N\,\!$. This implies $a^{n}\in N\,\!$, but this cannot be since $(a)\cap N=(e)\,\!$. So $o(aN)\geq o(a)\,\!$. 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 $|G|=p^{n}m\,\!$, 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 $p^{m}\,\!$