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Show by Lagrange's theorem that a group $G$ of order $27$ should have a subgroup of order $3$.

Attempt

We have $o(a)|o(G)$ for all $a\in G$. Then the possibilities of orders for the elements og $G$ are $1, 3, 9, 27$.

Let $e\neq x\in G$ then $x^{27}=e$ then $(x^9)^3=e$. Now let us consider a cyclic subgroup $H=<x^9>=\{e, x^9, x^{18}\}$. $H$ is a subgroup of $G$ of order $3$.

Is that proof error free? Is there any other method to solve the problem? Please use elementary tools (not use Sylow's theorem, Normal subgroup etc).

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  • $\begingroup$ Why is $x^9 \neq e$? $\endgroup$ – Aravind Oct 3 '15 at 6:22
  • $\begingroup$ How do you know that there is an element of order 3 though? If you can show existence of an element of order 3 then you are set. $\endgroup$ – HumbleStudent Oct 3 '15 at 6:22
  • $\begingroup$ Let $H=<x^t>$, where $t=o(x)/3$. $\endgroup$ – Aravind Oct 3 '15 at 6:32
  • $\begingroup$ @Aravind You have raised a right point. Then how to show that? If $x^9 \neq e$, then I think there is nothing more required? Am I right? $\endgroup$ – user1942348 Oct 3 '15 at 6:33
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Let $x\in G$ with $x\neq e$. By Lagrange theorem, the order of $x$ is $3$, $9$ or $27$.

  • If $x$ has order $3$, take $H = <x>$.
  • If $x$ has order $9$, then $x^3$ has order $9/3$. Take $H = <x^3>$.
  • If $x$ has order $27$, then $x^9$ has order $27/9$. Take $H = <x^9>$.
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  • $\begingroup$ How do you write "If $x$ has order $9$ and order $27$, then $x^3$ and $x^9$ have order $9/3$ and $27/9$ respectively. Please explain. $\endgroup$ – user1942348 Oct 3 '15 at 7:33
  • $\begingroup$ Let $G = <x>$ a cyclic group of order $n$ and $k > 0$ a divisor of $n$. Then $x^k$ has order $n/k$. Indeed, let $d$ the order of $x^k$. This is the smallest positive integer such that $(x^k)^d = e$, or equivantly, the smallest integer such that $n$ divides $kd$. As $k$ divides $n$, this is $d = n/k$. $\endgroup$ – Éric Guirbal Oct 3 '15 at 7:50
  • $\begingroup$ But here, G is not given as cyclic. Then $G = <x>$ is not meaningful. $\endgroup$ – user1942348 Oct 3 '15 at 9:38
  • $\begingroup$ I applied that theorem to the cyclic group $<x>$ where $x$ is any element of $G\setminus\{e\}$. For example, if $|<x>| = 9$, then $x^3$ is a subgroup of $<x>$ of order $9/3$ according to this theorem. $\endgroup$ – Éric Guirbal Oct 3 '15 at 9:59
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To see the error in your proof consider the case where the $x$ you chose has order $9$. You should be able to correct your proof though, as you are very close to a solution. Simply say: if $x$ does not have order $9$, then ... and if $x$ does have order $9$, then .....

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  • $\begingroup$ Would you please then write a full proof for the problem. I am understanding your point but not able to write the proof. $\endgroup$ – user1942348 Oct 3 '15 at 6:39
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Look, if the group is cyclic, then the existence of x in your proof is guaranteed, otherwise, such an x will not exist, so you need another case, one in which G is not cyclic. Case 2: Let us assume that no element of order 3 exists in G (which is non-cyclic) for if it does, then the cyclic group generated by that element will have order 3. So, all the elements of the group other than the identity, have order 9, but the number of elements of order d in a finite group is a multiple of phi(d), where phi is the Euler-Phi (Totient) Function. This is easy to prove, but now, this means that the number of elements of order 9 in G is a multiple of phi(9)=6, but since all the elements other than the identity are of order 9, there are 27-1=26 elements of order 9, so 6 divides 26, which is a contradiction, so there must be an element of order 3 in G, say p, so <(p)> is a subgroup of G of order 3.

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