group of order 27 must have a subgroup of order 3

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By Lagrange's theorem, the order of every element is a divisor of $27$.

Take any $g\in G$ with $g\ne1$.

If $ord(g)=27$, then $g^9$ has order $3$.

If $ord(g)=9$, then $g^3$ has order $3$.

If $ord(g)=3$, then $g$ has order $3$.

The same argument proves this:

If $p$ is prime, then every group of order $p^n$ has an element of order $p$.

Indeed, take any $g\in G$ with $g\ne1$. Let $ord(g)=p^m$. Then $g^{p^{m-1}}$ has order $p$.

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Updated on August 15, 2022

Comments

  • user304330
    user304330 about 1 year

    How to prove a group of order 27 must have a subgroup of order 3

    • Rory Daulton
      Rory Daulton almost 8 years
      This question was put on hold for being unclear. Just what is unclear about it? It looks perfectly straightforward to me. Perhaps this should be closed for showing no work on the part of the OP, but that is a different matter.
    • Rory Daulton
      Rory Daulton almost 8 years
      What work have you done on this problem, and just where are you stuck?
    • 6005
      6005 almost 8 years
      No reason to reopen it and close it again. Just leave it closed, it is very poor-quality.
  • user304330
    user304330 almost 8 years
    What is the meaning of " every group of order a power of p has an element of order p ." Please explain
  • Prince M
    Prince M over 7 years
    @user304330 ihf is just saying a more general proof that covers your question. Since your question was about a group of order 27 = $3^3$, its order is in the form $p^n$. Then what was shown to you was that the group must have an element of order $p$ as in $p^1$. And since the order of the element is the order of the subroutine generated by the element it covers your subgroup question.