# abstract algebra

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• November 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}
• November 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 $x^3 \equiv 1 \mod 17$ and $x^5 \equiv 1 \mod 17$ have the unique solution $x \equiv 1 \mod 17.$

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

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

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

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

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

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

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

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

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

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

Quote:

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