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G a group of order 35, H a normal subgroup of order 7. Prove that if g in G has order 7, then g is in H.

What I'm doing is invoking Sylow's third theorem to show that there exist a single subgroup of order 5 and a single subgroup of order 7. And then the direct product of these has order 35, and then somehow... this means G has an element of order 35, which would mean G is cyclic, and if G is cyclic, each of its subgroups are cyclic. So H has an element of order 7 and if there's an element g in G of order 7, then it generates the subgroup H?

This is my train of logic? I could use some help on the somehow... part, or if that's even a valid way of proving that a group of order 35 is cyclic. Please help steer me back on track.

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3 Answers 3

up vote 3 down vote accepted

Since $|G|=35=7\cdot 5$, a subgroup $H$ of order $7$ is a Sylow-$7$-subgroup. Moreover, given that $H$ is normal, it must be the only Sylow-$7$-subgroup of $G$. Now, if $g\in G$ has order $7$, then $\langle g\rangle $ is a $7$-subgroup of $G$ and hence is contained in a Sylow-$7$-subgroup of $G$. But $H$ is the only Sylow-$7$-subgroup, so $\langle g\rangle\subseteq H$. Therefore, $g\in H$.

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@Spenser: ${+3+(-1)+(+1)=3}$ –  Babak S. Nov 26 '13 at 17:59
    
@B.S. That is very nice of you ! Love this site ! Thank you ! :) –  Spenser Nov 26 '13 at 19:15

Let $K=\langle g\rangle$ be a subgroup of $G$ of order $7$. Since $H$ is a normal so $HK\leq G$ and so $|HK|\bigg||G|$. This means that $$\frac{7\times 7}{s}\mid 7\times 5$$ in which $|H\cap K|=s$. But $s=1$ cannot be happen.

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Very nice solution! +1 –  Spenser Nov 26 '13 at 17:17

An element of order 7 has to be in a Sylow 7-subgroup. But all the Sylow 7-subgroups are conjugate, $H$ is one of them, and, being normal, $H$ has no conjugates other than itself.

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