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Let H be a normal subgroup of a finite group $G$ and let $x \in G$. If $\gcd\left ( \left | x \right |,\left | G/H \right | \right )=1$, show that $x \in H$.

Attempt:

Let $H \triangleleft G$. So, $\forall x \in G: xH =Hx \Rightarrow xHx\subseteq H$

Since H is a normal subgroup of G: $\left | G:H \right |=\frac{\left | G \right |}{\left | H \right |}=\left | \frac{G}{H} \right |=2$.

I would appreciate hints to keep me going.

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  • $\begingroup$ Why is $|G/H|=2$ again? $\endgroup$ – RKD Jul 10 '17 at 4:08
  • $\begingroup$ Because the index of a normal subgroup is 2. $\endgroup$ – Mathematicing Jul 10 '17 at 4:08
  • $\begingroup$ This is not true. Normal subgroups can be of arbitrary index. For example, in an Abelian group, every subgroup is normal. Are you saying then that every subgroup has index 2? $\endgroup$ – RKD Jul 10 '17 at 4:09
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Note: Your conclusion that $\because H$ is a normal subgroup of $G$, hence $[G:H]=2$ is incorrect (as noted in comments as well). All we can say is that $[G:H]$ is finite (because $|G|$ is finite).

Let $|x|=a$ and $|G/H|=b$. We are given that $\gcd(a,b)=1$. Therefore there exists $m,n \in \mathbb{Z}$ such that $ma+nb=1$.

Now consider the coset $xH$. $$(xH)^a=x^aH=eH=H \quad \implies \quad x^a \in H.$$ Likewise $$x^bH=(xH)^b=H \quad \implies \quad x^b \in H.$$ Now $$x=x^1=x^{ma+nb}=\underbrace{x^{ma}}_{\in H} \, \overbrace{x^{nb}}^{\in H} \in H$$

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Hint: Suppose $x\notin H$, then look at the coset $xH$ in the group $G/H$. What does Lagrange's theorem say about the order of $xH$?

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  • $\begingroup$ The order of xH divides the order of G/H but the order of xH is just the order of x.... $\endgroup$ – Mathematicing Jul 10 '17 at 4:13
  • $\begingroup$ Well the order of $xH$ doesn't have to be equal to the order of $x$, but it does need to divide it. $\endgroup$ – RKD Jul 10 '17 at 4:15
  • $\begingroup$ If the order of xH divides the order of x, then the order of x is some multiple of the order of xH $\endgroup$ – Mathematicing Jul 10 '17 at 4:21
  • $\begingroup$ Yes, now conclude using the gcd assumption. $\endgroup$ – RKD Jul 10 '17 at 4:22
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    $\begingroup$ The point is that the order of $xH$ divides $|x|$ and $|G/H|$ so it has to divide their gcd, which is $1$. so $x\in H$, a contradiction. $\endgroup$ – RKD Jul 10 '17 at 4:54

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