1. ## abelian normal group

Let $G$ be a group of order $3825$. Prove that if $H$ is a normal subgroup of order $17$ in $G$ then $H \leq Z(G)$, where $Z(G)$ represents the center of $G$.
(Please do not use sylow's or cauchy's theorems.)

2. Originally Posted by abhishekkgp
Let $G$ be a group of order $3825$. Prove that if $H$ is a normal subgroup of order $17$ in $G$ then $H \leq Z(G)$, where $Z(G)$ represents the center of $G$.
(Please do not use sylow's or cauchy's theorems.)

If you don't want Sylow and Cauchy theorems, then I suppose you want even less semidirect

products and stuff, thus the question is: what do you want? What stuff is allowed or wanted by you?

Tonio

3. Is there a nice way to do this if the Sylow theorems are allowed? I can't see anything short of a group classification argument.

4. Originally Posted by roninpro
Is there a nice way to do this if the Sylow theorems are allowed? I can't see anything short of a group classification argument.
Indeed, I feel as though the simplest way to do this is to note that it is of the form $p^2q$ with $p,q$ prime and $q\notequiv 1\text{ mod }p$ and thus abelian...that said the only ways I know how to prove this are Sylow's theorems or rep. theory.

5. Just noting the arithmetic: $3825=3^2\cdot 5^2\cdot 17$.

6. Originally Posted by abhishekkgp
Let $G$ be a group of order $3825$. Prove that if $H$ is a normal subgroup of order $17$ in $G$ then $H \leq Z(G)$, where $Z(G)$ represents the center of $G$.
(Please do not use sylow's or cauchy's theorems.)
(*) If H is a subgroup of G, then the factor group $N_G(H)/C_G(H)$ is isomorphic to a subgroup of Aut H (Hungerford, "Algebra", p92).

By assumption, $N_G(H)=G$. Since H is a group of order 17, H is isomorphic to the cyclic group of order 17, i.e., $C_{17}$. Thus, |Aut(H)|=16.

Now, $|G/C_G(H)|$ divides both 16 and 3825 by Lagrange's theorem and (*), it follows that $|G/C_G(H)|=1$. Can you conclude from here?