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Let $G$ be a group, and $N \lhd G$ so that $[G:N]=2014$.
I need to prove that there is a subgroup of index 2.

Indeed, since $N \lhd G$, $o(G/N)=2014$, but I couldn't find how to go on.

Ran Kashtan
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3 Answers3

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Work in $G/N$:

We have $2014=2\times 19\times 53$.

Hence there exist a subgroup of order $53$. This subgroup is either normal, or there would be at least $54$ such subgroup. Since $53$ is a prime, any 2 subgroup of order $53$ must intersect only at $e$. So there cannot be $54$ subgroup of order $53$. Thus the subgroup of order $53$ is a normal subgroup $H$. Hence we quotient that one out again. And the rest is easy.

Then the rest is just standard correspondence theorem.

Gina
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  • Maybe you meant that there would be at least 38 such subgroup (instead of 54)? – Ran Kashtan Jan 03 '14 at 10:55
  • I'm not sure where the $38$ come from, but no: $38\not\equiv 1(\mod 53)$ – Gina Jan 03 '14 at 11:01
  • I have 1 normal subgroup of order 53, and 1 normal subgroup of order 19. I'm not sure how to go on.. – Ran Kashtan Jan 03 '14 at 12:59
  • If both group are normal, then just take their product. (actually, you just need 1 normal group) – Gina Jan 03 '14 at 13:04
  • Can you please explain more how to go on? – Ran Kashtan Jan 03 '14 at 13:18
  • Let's say $H$ and $K$ is the group of order $53$ and $19$, in which $H$ is normal. Then $HK$ is a group, and since $H$ and $K$ only intersect at $e$ the order is $53\times 19$. That's one way. Or you can just take the quotient $(G/N)/H$ and find another subgroup in that and then apply correspondence twice. You might want to add in to your question exactly how far have you got in group theory, because it's possible that you still got stuck because people are referring to theorem you have not learnt. – Gina Jan 03 '14 at 13:25
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Hint: use the Sylow-Theorems to show that the group $G/N$ of order $2014=2\cdot 19 \cdot 53$ has a subgroup $H/N$ of index $2$. Then use the Correspondence Theorem to conclude that this corresponds to $H<G$ with $(G:H)=2$.

benh
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Obviously, we can just replace $G$ with $G/N$, and assume that $G$ itself has order 2014.

We have $2014 = 2 \times 19 \times 53$.

The number of $53$-Sylow subgroups of $G$ divides $2 \times 19 = 38$, but at the same time is congruent to $1 \mod 53$. So there is only one subgroup of $G$ of order 53. Therefore that subgroup is normal. (Any conjugate of it can only be itself.)

So modding out by that subgroup, we can now assume that $G$ has order 38. A $19$-Sylow subgroup will do the trick.

user118829
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