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This was a previously asked question that only had answers referring to Sylow's theorem, which is not something I can use here.

Let $G$ be a group, $|G|=35.$ We know that $G$ must contain an element of order $5$ and an element of order $7$.

(1) Prove that $G$ must have a normal subgroup of order $5$ or of order $7$ or both.

(2) Deduce that $G$ is isomorphic to $C_{35}$.

(1) Attempt: It is shown earlier in the question that if $\Omega$ is the set of subgroups of $G$ isomorphic to a subgroup $H\leq G,$ and $G$ acts on $\Omega$ by conjugation, then $$\text{Stab}(H)=G \iff H\; \text{is a normal subgroup}.(*)$$

Assume for contradiction $G$ has no normal subgroups of order $5$ or $7$. Take $g_1,g_2$ as order $5$ and order $7$ respectively, then $\langle g_1\rangle$,$\langle g_2\rangle \leq G$ are not normal.

Now by $(*)$ we can find an element in $G$ such that conjugating $\langle g_1\rangle$ by it gives a subgroup of $G$, isomorphic to but distinct from $\langle g_1\rangle$. By Lagrange's theorem then, and noting the intersection of subgroups is a subgroup, these two order $5$ subgroups must have trivial intersection

Similarly we can find two order $7$ subgroups with trivial intersection. And it is clear too that all non-identity elements in all four of these subgroups, of which there are $20$, are distinct.

This leaves $14$ non-identity elements outside of these subgroups, but I am stuck here.

I think if $G$ has no order $35$ element then noting the only non-negative integer solution to $4p+6q=14$ is $p=2,q=1$ shows that we can distribute these elements into two new distinct order $5$ subgroups and one new order $7$ subgroup, but I can't see how this would help.

I'm sure I can complete this as a counting argument but I'm not sure where to go.

Shaun
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jcneek
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    All caps is the internet equivalent of yelling, not of emphasis. – Arturo Magidin May 11 '21 at 21:39
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    The normalizer of the subgroup of order $7$ must have order $7$ or $35$, by Lagrange. If it has order $35$, you are done. If it has order $7$, that means that there are $5$ conjugates, any two of which intersect trivially, giving you $31$ elements of order dividing $7$, and leaving only four potential elements to make a subgroup of order $5$. – Arturo Magidin May 11 '21 at 21:41
  • Sorry why exactly must the normaliser be order $7$ or $35$? Is it because it must be a subgroup itself, and if so why is this true? – jcneek May 11 '21 at 22:06
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    Yes, the normalizer of a subgroup $H$ is a subgroup of $G$ that contains $H$. If $S$ is the subgroup of order $7$, and $N$ is its normalizer, then $S\leq N\leq G$, so $[G:N][N:S]=5$. The only possibilities are $[G:N]=5$ and $N=S$, or $[N:S]=5$ and $G=N$. The normalizer is what you called the stabilizer. – Arturo Magidin May 11 '21 at 22:09
  • Ah OK I'm with you. So in the circumstance that the normaliser is order $7$, should I then be able to show that the remaining $4$ elements with $e$ give a normal subgroup? – jcneek May 11 '21 at 22:21
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    When the normalizer is order $7$, there are only $4$ elements left not included in any subgroup of order $7$. Those four elements and $e$ are then the only possible subgroup of order $5$, and if there is a unique subgroup of order $5$, then that subgroup is normal. – Arturo Magidin May 11 '21 at 22:22
  • Great I understand thank you. Do you know how best to approach the second part of the question? – jcneek May 11 '21 at 22:25
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    Show the conjugation action of the other subgroup on the normal subgroup must be trivial, because there aren’t any automorphisms of the “right” order for the action to be nontrivial. Conclude the two subgroups commute elementwise. – Arturo Magidin May 11 '21 at 22:26

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