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Is it true or not? the question doesn't mention whether G is abelian or cyclic. It just say it is a finite group

somebody told me it is false,but I cannot find a counter-example.

But, for me I would say it is true

as By Lagrange Theorem, for any finite group G, any subgroup H<=G, |H| is always a divisor of |G|.

Therefore, if d is a divisor of G, then d should be the same as |H|(?)

Ben Ho
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    Please edit to include your efforts. – lulu Oct 29 '24 at 10:41
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    I believe $A_4$ is a counterexample. – Jean Oct 29 '24 at 10:49
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    I assume this is a question from an intro course. Depending on how far along this course you are, there is nothing better than to look for small counterexamples, $A_4$ being the smallest such one.

    An interesting question is why this statement shouldn't hold, i.e. how you should get the impression to look for counterexamples. Maybe you've seen Cauchy's Theorem or the first Sylow Theorem that are strong statements about the existence of certain subgrouups. Maybe that should give you the hint that existence of certain subgroups is something special and not something that "goes without saying".

     – SometimesBlind Oct 29 '24 at 10:52
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    The fact that $A\implies B$ does not generally imply that $B\implies A$. – lulu Oct 29 '24 at 10:56
  • No, you can build counter-example. This quetion can be related to Lagrange’s theorem : https://www.youtube.com/watch?v=TCcSZEL_3CQ&t=72 – leofun01 Oct 29 '24 at 11:06
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    As mentioned above, this is false. But I just thought I'd mention that it is true when d is a power of a prime (i.e. $d=p^k$ for some prime $p$). This is a famous result known as the first Sylow Theorem. Also, the result is true when the group is cyclic, which is a consequence of what is generally referred to as the Fundamental Theorem of cyclic groups. – David Reed Oct 29 '24 at 11:07
  • Sylow doesn't say this for every $k,$ does it? Been a long time, but I thought it only did it for maximal $k.$ Tht is, $p^k\mid n$ but not $p^{k=+1}.3 @DavidReed – Thomas Andrews Oct 29 '24 at 11:29
  • It is, however, true for abelian groups. But, for a counter example, I bet there is no subgroup of $S_5$ of order $15.$ There is not one group (up to isomorphism) of order $15,$ the cyclic group, and it is definitely not in $S_5.$ – Thomas Andrews Oct 29 '24 at 11:32
  • More than you want to know at https://math.stackexchange.com/questions/2144077/complete-classification-of-the-groups-for-which-converse-of-lagranges-theorem-h – Gerry Myerson Oct 29 '24 at 12:19
  • @ThomasAndrews It is for every k, although the other sylow theorems focus on the maximal power. See here: https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Abstract_Algebra%3A_Theory_and_Applications_(Judson)/15%3A_The_Sylow_Theorems/15.01%3A_The_Sylow_Theorems – David Reed Oct 29 '24 at 12:47

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