# abstract algebra

• Nov 3rd 2008, 06:07 PM
dori1123
abstract algebra
Let G be a group of order 3825. Prove that if H is a normal subgroup of order 17 in G, then H is a subgroup of Z(G).

Z(G) = { g in G | xg = gx for every x in G}
• Nov 4th 2008, 04:48 AM
NonCommAlg
Quote:

Originally Posted by dori1123
Let G be a group of order 3825. Prove that if H is a normal subgroup of order 17 in G, then H is a subgroup of Z(G).

Z(G) = { g in G | xg = gx for every x in G}

Lemma: both equations $\displaystyle x^3 \equiv 1 \mod 17$ and $\displaystyle x^5 \equiv 1 \mod 17$ have the unique solution $\displaystyle x \equiv 1 \mod 17.$

Proof: if $\displaystyle x^3 \equiv 1 \mod 17,$ then $\displaystyle x^{15} \equiv 1 \mod 17.$ hence $\displaystyle x^{16} \equiv x \mod 17.$ but $\displaystyle x^{16} \equiv 1 \mod 17.$ thus: $\displaystyle x \equiv 1 \mod 17.$ a similar argument works for $\displaystyle x^5 \equiv 1 \mod 17. \ \ \ \Box$

now suppose $\displaystyle |G|=3825=3^2 \times 5^2 \times 17$ and let $\displaystyle H=<h>$ be the normal subgroup of G with $\displaystyle o(h)=17.$ in order to show that $\displaystyle H \subseteq Z(G)$, we only need to prove that $\displaystyle h \in Z(G)$.

let $\displaystyle g \in G.$ we want to show that $\displaystyle g$ commutes with $\displaystyle h.$ let $\displaystyle o(g)=m.$ since H is normal, there exists an integer $\displaystyle n$ such that $\displaystyle ghg^{-1}=h^n.$ thus: $\displaystyle h=g^mhg^{-m}=h^{n^m},$ which gives us

$\displaystyle n^m \equiv 1 \mod 17.$ we have $\displaystyle m=3^r \times 5^s \times 17^t,$ for some $\displaystyle 0 \leq r \leq 2, \ 0 \leq s \leq 2, \ 0 \leq t \leq 1.$ since $\displaystyle n^{17} \equiv n \mod 17,$ we may assume that $\displaystyle t=0.$ now by the Lemma we must have

$\displaystyle n \equiv 1 \mod 17$ and thus $\displaystyle ghg^{-1}=h^n=h.$ therefore $\displaystyle gh=hg,$ i.e. $\displaystyle h \in Z(G). \ \ \Box$

Exercise: (very easy!) make the problem interesting by extending this result to a class of finite groups!
• Nov 4th 2008, 12:55 PM
ThePerfectHacker
Quote:

Originally Posted by NonCommAlg
Lemma: both equations $\displaystyle x^3 \equiv 1 \mod 17$ and $\displaystyle x^5 \equiv 1 \mod 17$ have the unique solutions $\displaystyle x \equiv 1 \mod 17.$

Proof: if $\displaystyle x^3 \equiv 1 \mod 17,$ then $\displaystyle x^{15} \equiv 1 \mod 17.$ hence $\displaystyle x^{16} \equiv x \mod 17.$ but $\displaystyle x^{16} \equiv 1 \mod 17.$ thus: $\displaystyle x \equiv 1 \mod 17.$ a similar argument works for $\displaystyle x^5 \equiv 1 \mod 17. \ \ \ \Box$

Here is another way (not that there is anything bad about what you did) for dori.

If you have the equation $\displaystyle x^q \equiv 1(\bmod p)$ and $\displaystyle q\not | (p-1)$ then $\displaystyle x\equiv 1(\bmod p)$ where $\displaystyle p,q$ are primes. This is because $\displaystyle \mathbb{Z}_p^{\times}$ is a group. And if $\displaystyle x^q \equiv 1$ and $\displaystyle x\not \equiv 1$ then it means $\displaystyle x$ has order $\displaystyle q$. By Lagrange's theorem it means $\displaystyle q$ divides $\displaystyle |\mathbb{Z}_p^{\times}| = p-1$ a contradiction.

Quote:

thus $\displaystyle ghg^{-1}=h^n=h.$ therefore $\displaystyle xh=hx,$ i.e. $\displaystyle h \in Z(G). \ \ \Box$
You mean $\displaystyle gh=hg$. But whatever it is not any significant.