recently my teacher has introduced us two new Theorems called the N/C Theorem and the Normalizer Theorem.
N/C states: Let $H<G$ and $N(H)$ be a normalizer of H in G. Let $C(H)$ be the centralizer of H. Then $\phi:N(H)\rightarrow Aut(H)$ is a homomorphism with kernal $\frac{N(H)}{C(H)}$ is isomorphic to $\phi(N(H))<Aut(H)$ The mapping is $\phi(g)=\gamma_g$ where $\gamma_g(x)=gxg^{-1}$
Normalizer Thm: Let $H<G$ and N(H) be the normalizer. Then the normalizer is a subgroup of G and H is normal to N(H).
Now he asks us a question where I am assuming you have to use those theorems to help.
The question is:
Let G be a group of odd order with less than 200 elements with an element h
of order 23. What are the possible order of G and prove H=<h> must be a normal
subgroup of G.
Since G is of odd order and h is in G, then all possible order of G are $23, 23*3=69, 23*5=115, 23*7=161$. Right? Then to show that H is normal would there be a way to show that G is cyclic, thus every subgroup of it is normal?
The question also asks
a) Why does C(H) contain H?
b) Use the N/C Theorem to show why |G/C(H)|=1. Conclude that h commutes with all elements in G.
c) Show that G/H must be cyclic and use the fact that $H\subset Z(G)$ to show that G must be abelian.
a) This is because for modulo arithmetic is commutative. So for every g in G, gh=hg for every h in H.
b)This is where I would like the most help since I dont know where to begin.
c) I feel like there is a Thm about this, but this is what shows G is abelian so I am already second guessing my idea for the original part of my explanation.
Best Answer
Proposition Let $G$ be a group of odd order with less than $200$ elements and let $23$ divide the order of $G$. Then $G \cong C_{23}, C_{69}, C_{115}$ or $C_{161}$.
Proof By Cauchy’s Theorem we can find an element $h \in G$ of order $23$. Put $H=\langle h \rangle$. Owing to the $N/C-Theorem$, $N_G(H)/C_G(H)$ can be homomorphically embedded in Aut$(H) \cong C_{23}^* \cong C_{22}$. Since $|G|$ is odd, it follows that $|N_G(H)/C_G(H)|$ must divide $11$. If $|N_G(H)/C_G(H)|=11$, then, since $H \subseteq C_G(H)$, we would have $$|G| \geq\text{ index }[N_G(H):C_G(H)] \cdot \text {index}[C_G(H):H] \cdot |H| \geq 11\cdot 23 = 253,$$contradicting the fact that the order of $G$ is smaller than $200$. We conclude that $ N_G(H)=C_G(H)$.
Observe that $23^2=529 \nmid |G|$, hence $H \in Syl_{23}(G)$. Assume that $\#Syl_{23}(G) \gt 1$. Then by Sylow’s Theorems, we would have $\#Syl_{23}(G)=\text {index}[G:N_G(H)] \geq 24$. But then $|G| \geq 23 \cdot 24 = 552$, again contradicting $G$ having less than $200$ elements. We conclude that index$[G:N_G(H)]=1$, so $G=N_G(H)=C_G(H)$, meaning $H \subseteq Z(G)$.
Finally observe that $|G/Z(G)| \mid |G/H|$. Since $|G|$ is odd, less than $200$ and $|H|=23$, $|G/Z(G)| \in \{1,3,5,7\}$. This means that $G/Z(G)$ is cyclic, and it is well known that in this case $G$ must be abelian. The proposition now follows easily.