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I want to verify whether there exist subgroups of $S_5$ of order $15,20,40$. but I don't know how to approach. generally, how do you prove when you are asked to find a subgroup with specific order of a finite group? any good/efficient way other than brute forcing with random generators?

user159234
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    I wouldn't use "random" generators. I'd try generators whose orders as individual elements divide the order of the desired subgroup. Playing around with those, you'll probably start to learn how they interact. – G Tony Jacobs Jun 13 '17 at 15:00
  • In a subgroup of $S_5$ of order $15,20$ or $40$ there is a $5$-cycle by Cauchy's theorem. If the order is $15$ there also is a $3$-cycle. This does not leave much to check. – Jack D'Aurizio Jun 13 '17 at 15:00
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    "Every subgroup of $S_n$ is either all even or half-even and half odd." This is usually very handy when dealing with $S_n$. – Sahiba Arora Jun 13 '17 at 15:07
  • @JackD'Aurizio how do you know there is a $3$-cycle? is there no element other than $3$-cycle having order $3$? – user159234 Jun 13 '17 at 15:25
  • @user159234: $3+3>5$ hence any order-$3$ element in $S_5$ is a $3$-cycle. – Jack D'Aurizio Jun 13 '17 at 15:31

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If it is just for verification and not for the exam/assignment, then you can also verify using computer. You can run the following GAP code:

G := SymmetricGroup( 5 );
List( AllSubgroups( G ), i -> Order( i ) );

After this you can visually verify, whether there is any subgroup of Order 15,20 and 40.

To get the actual subgroups (of order 20 in your case), you can use:

Filtered( l, i -> Order( i ) = 20 )

Hope this helps!

-- Mike

Mike V.D.C.
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  • Please note however that AllSubgroups was intended for teaching purposes for small groups. See GAP F.A.Q. for some hints. – Olexandr Konovalov Jun 13 '17 at 21:24
  • @AlexanderKonovalov, Agreed and noted. However in this example, as $S_5$ is small enough, so we can even verify the list or orders. – Mike V.D.C. Jun 13 '17 at 21:53
  • Yes, it certainly works here, and it may be useful for teaching purposes (when GAP is uses to teach algebra) because it reduces cognitive load. AllSubgroups is very often advised on this site, but important also to tell that this is not the only way of doing this, and not the most efficient one. I've once posted this answer to give some hints. – Olexandr Konovalov Jun 13 '17 at 22:08
  • @AlexanderKonovalov, thanks for the explanation and especially the link. – Mike V.D.C. Jun 14 '17 at 06:01
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$S_5$ has no subgroup of order $40$, see this duplicate for the arguments. The same holds for order $15$. For a subgroup of order $20$ we can take an element of order $5$, which exists by Cauchy, i.e., a $5$-cycle $(12345)$ and a $4$-cycle $(2354)$ to obtain a subgroup of order $20$. Here the $4$-cycle normalizes the subgroup generated by $(12345)$. In fact, the normalizer $N(P)$ of an order $5$ subgroup $P$ of $S_5$ has $20$ elements.

Dietrich Burde
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  • do you actually compute all 20 elements? or is there any method to compute the order of normalizer? – user159234 Jun 13 '17 at 15:14
  • We already know the order without counting. If $n$ is the order, then $4\mid n$ and $5\mid n$ by Lagrange. We can also argue that $n\le 20$. – Dietrich Burde Jun 13 '17 at 18:45