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I was given the following question: Does every group of order 56 contain a subgroup of order 2. I know that the Sylow theorems guarantee the existence of an order 8 subgroup. Is there a general technique to prove the existence or the non-existence of subgroups of lower prime powers? Thank you for your help.

mathology
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    There is a more elementary proof that every finite group of even order has an element of order 2, which does not use Cauchy's theorem or Sylow's theorem. Just pair every element off with its inverse, and notice that at least two elements in the group must be equal to their own inverse. – Derek Holt Jun 20 '19 at 10:15

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Let $G$ be a group of order $56$. If $g\in G$ has even order $2n$, then the order of $g^n$ is $2$, and you are done. What if every element has odd order? Since the order if each element must divide $56$, that would mean that every $g\in G\setminus\{e\}$ must have order $7$. But the intersection of any two distinct subgroups of order $7$ is $\{e\}$. So, since $G$ coud be written has the union of subgroups of order $7$ and since any two such subgroups, if they are distinct, only share $e$ as a common element, $56$ could be written has $1+6k$, for some $k\in\mathbb N$. However, no such $k$ exists.

José Carlos Santos
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  • This is still more complicated than the elementary proof that finite groups of even order have an element of order $2$. – Derek Holt Jun 20 '19 at 12:19
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Hint : It is well known that if a prime $p$ divides the order of a group $G$ , an element of order $p$ in $G$ exists. Take it from here.

Peter
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    Thank you for your help. Now it is all clear. Unfortunately, we did not cover Cauchy's theorem in class. – mathology Jun 20 '19 at 09:26
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As you say, the first Sylow theorem tells you that there is a subgroup of order $8$. Take a non-identity element $g$ of that subgroup. It has either order $2, 4$ or $8$, by Lagrange's theorem. If $g$ has order $2$, we're done. If $g$ has order $4$, then $g^2$ has order $2$, and we're done. If $g$ has order $8$, then $g^4$ has order $2$, and we're done.

Arthur
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  • Thank you for your answer. Is this already a proof of Cauchy's theorem? – mathology Jun 20 '19 at 09:29
  • @PhilippWeder I would personally consider Cauchy's theorem more basic than Sylow's, but if you've learned Sylow's and not Cauchy's, then yes, this generalizes neatly to a proof of Cauchy's. – Arthur Jun 20 '19 at 09:31
  • Oh nice, thank you! I mostly used Basic Algebra I by Nathan Jacobson and he only shows a special case of Cauchy's theorem in the proof of the Sylow theorem, but he does not cover the general statement. – mathology Jun 20 '19 at 09:33
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A well-known fact from group theory is that a group of order $p^n$ has subgroups of order $p^m$ for all $0\leq m\leq n$.

Proof. By induction on $n$. A non trivial subgroup has non trivial center. Pick $z$ in the center, apply induction to $G/\langle z\rangle$, and lift your subgroups to $G$ (that is take their inverse images by the canonical projection).

Applying this result to Sylow subgroups gives you the following theorem.

If a finite group $G$ has order divisible by $p^n$, it has subgroups of order $p^m,$ for all $ 0\leq m\leq n$.

GreginGre
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