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Question Let $G$ be a group, $|G|=35.$ We know that $G$ must contain an element of order $5$ and an element of order $7$. Prove that $G$ must then have a normal subgroup of order $5$ or of order $7$ or both.

Attempt. Suppose for a contradiction that $G$ does not have a normal subgroup of order $5$ nor of order $7$. Then there are at least two subgroups of order $5$ and at least two subgroups of order $7$. This is because conjugation of a subgroup $g^{-1}Hg$ is still a subgroup. But we see that subgroups of order $5$ or $7$ must either be disjoint (with the exception of the identity $e$) because intersection of two subgroups is still a subgroup. Also, by our hypothesis, $G$ does not contain an element of order $35$.

How to continue from here?

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  • $\begingroup$ For clarification, this appears in an introductory group theory course, so Sylow Theorem should not be used. $\endgroup$
    – Bernoulli
    Apr 9, 2018 at 20:11

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Maybe look into Sylow's theorem as I will for subgroup of order $5$: let $n_5$ be the number of Sylow $5$-subgroups of $G$ (note $5^2 \nmid 35$ ).

Then $n_5$ divides 7 and also $n_5 \equiv 1 \mod 5 \Rightarrow n_5 = 1$ and so the subgroup of order $5$ is unique, and therefore it must be normal.

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Hint:
Apply Sylow's theorem. Let $n_p$ be the number of Sylow p-subgroups in G, where p=7,5. Recall that if $n_p$=1, then the unique Sylow p-subgroup is normal in G.

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