Normal subgroup in center of the group 
Let $G$ be a group of order $3825$. Prove that if $H$ is a normal subgroup of order $17$ in $G$ then $H\leq Z(G)$.  

In the link below, the solution basically says that the index of $C_{G}(H)$ must divide 225, and $Z(G)=17-1=16$ and so the index is $1, $ and $G=C_{G}(H).$ Why can't the index be 3, 5, or 15?
http://crazyproject.wordpress.com/2010/05/20/if-a-subgroup-of-order-17-is-normal-in-a-group-of-order-3825-then-it-is-central/
 A: If $H$ is normal then $G$ acts on $H$ by conjugation, inducing a homomorphism $G\to{\rm Aut}(H)$ with kernel $C_G(H)$. Therefore $G/C_G(H)\le{\rm Aut}(H)$, as is stated in the link. Thus, the index of $C_G(H)$ must divide $|{\rm Aut}(H)|=|U(17)|=\varphi(17)=16$ by Lagrange's theorem. Since $H$ is cyclic it is abelian and so $H\le C_G(H)$, hence the index of $C_G(H)$ divides the index of $H$, which is $3^25^2$. An integer dividing $16$ and $3^25^2$ must divide $\gcd(16,3^25^2)=1$, so the index is $1$ and $C_G(H)=G$.
A: 
THEOREMS USED-



*

***Let $H  $ be a normal subgroup of a group $G$. Then $G $ acts by conjugation on $H$ as automorphisms of $H$.More specifically,the action of $G$ on $H$ by conjugation by $G$ is defined for each $g$$\in$$G$ by $$h\mapsto ghg^{-1}$$ $\forall  h\in H$


For each $g\in G$,conjugation by $g$ is an automorphism of $H$.The permutation representation afforded by this action is a homomorphism of $G$ into $Aut(H)$ with kernel $C_G(H)$.In particular, $G/C_G(H)$ is isomporphic to  a subgroup of $Aut(H)$.**


*

*The automorphism group of a cyclic group of order n is isomorphic to U(n),an abelian group of order φ(n),Where φ is an Euler's function.

*Lagrange's theorem-If $G$ is a finite group and $H$ is a subgroup of $G$,then the order of $H$ divides the order of $G$.

SOLUTION:

If  $H\unlhd G$,then $G$ acts by conjugation on $H$ as automorphism of H,with kernel $C_G(H)$
Hence,$G/C_G(H)\le{\rm Aut}(H)$.Since,H is of prime(17) order.So,$H$ is cyclic.So,$|{\rm Aut}(H)|=\varphi(17)=17-1=16$ .
Since,$G/C_G(H)\le{\rm Aut}(H)$,by Lagrange's theorem $|G/C_G(H)|$ divides $|{\rm Aut}(H)|$ this implies possible orders for $|G/C_G(H)|$ are $1,2,4,8.$-----(*)
Also,$C_G(H)\le{\rm G}$, again by Lagrange's theorem $|C_G(H)|$ divides $|G|$. 
From (*), $|C_G(H)|$=$3825/2$ or $3825/4$ 0r $3825/8$
.But none of,$3825/2$ , $3825/4$ , $3825/8$$\in$$\mathbb Z$.Then only possibility left is $|C_G(H)|$=$3825/1$=$3825$=$|G|$ implying $G=C_G(H)$.
Since,$H$ is abelian ,$H\le{\rm C_G}(H)$ & $C_G(H)\le{\rm Z}(G)$.Thus,$H\le{\rm Z}(G)$
