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I'm preparing for an exam, and this little guy just had me stumped:

Let $G$ be a group, $N\subset G$ a normal subgroup of $G$, $x,y,z \in G$, and $x^3 \in N$, $y^5 \in N$, $zxz^{-1}y^{-1} \in N$. Show that $x,y\in N$.

What I want to do is to show for that for $g \in G$ that $gxg^{-1} \in N$. Another approach I tried is to just multiply the 3 given elements in $N$ in certain ways and use that $N$ is closed under multiplication to show that $x \in N$. No luck so far. This can't be that difficult right?

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Look in the quotient group $G/N$. It appears that $x$ has order 3, and $y$ has order 5, but $x$ and $y$ are conjugate by $z$, so they have the same order, $1$. – Jack Schmidt Aug 7 '12 at 21:31
Where did you get that $x$ is order 3 and $y$ is order 5? – BallzofFury Aug 7 '12 at 21:50
@BallzofFury: $\,x^3\in N\Longrightarrow (xN)^3=N\,$ in the quotient $\,G/N\,$ . So it appears the order of $\,xN\,$ is $\,3\,$ but, in fact, it is one. You can read my answer below, too. – DonAntonio Aug 7 '12 at 21:55
up vote 3 down vote accepted

Adding on Jack's hint (well, in fact solution!): we have in the quotient $$G/N:\,\,\,(xN)^3\in N\,\,,\,(yN)^5\in N\,\,,\,(zxz^{-1}N)=(ynN)=yN\Longrightarrow$$ the elements $\,xN\,,\,yN\,\in G/N\,$ are conjugated (by $\,zN\,$) and thus they have the same order...but $\,(3,5)=1\,$ , so it must be that $$xN=yN=N=\,\text{the unit element in the quotient}\,\Longleftrightarrow x\,,\,y\in N$$

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Beautiful and elegant. Hadn't thought of working like that in the quotient group. Many thanks! – BallzofFury Aug 8 '12 at 8:28

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