This is what I have. Note: I'm not allowed Cauchy's theorem or Sylow theorems.

Let $|G| = 275$. So I know $275 = 5\times5\times11$. If I assume that $G$ is cyclic then there exists $x\in G$ such that $|x| = 275$. Then $|x^{55}| = 5$ and I'm done.

If $G$ is not cyclic then I assume it contains no elements of order $275$. Then all elements of $G$ have order $1,5,11,25,55$. It has an element of order $5$ then I stop because we're done. So I look at an element of order $|x| = 25 \text{ or } 55$ but then wouldn't $|x^{5 \text{ or } 11}| = 5$ so I'd be done here also. Thus I can say $G$ has elements of order $11$ only. So for each element of order $11$ I can look at $\langle x\rangle$ and this will have $|\langle x\rangle|=11$ for each distinct such subgroup I'll get $10$ distinct elements and so if there are k such subgroups I'll have $10k + 1$ elements and $275 = 10k+1 \iff 274 = 10k$ but $10$ doesn't divide $274$ so $G$ must have an element of order $5$.

Is this enough or do I need to show more and perhaps more importantly does this work?

(edit: spelling)

  • 6
    $\begingroup$ Seems fine to me. $\endgroup$ – lhf Nov 5 '13 at 1:23
  • $\begingroup$ This is a direct result of the Cauchy's theorem or Sylow's theorem in case you have studied them. $\endgroup$ – user53970 Nov 5 '13 at 1:29
  • $\begingroup$ @user53970 , I know of them through independent study but they've not appeared in the course so I don't think I'n allowed to use them. $\endgroup$ – AvatarOfChronos Nov 5 '13 at 1:33
  • $\begingroup$ Your argument is correct. The only thing I think you should add is why $\langle x \rangle \cap \langle y \rangle = e $ for all $x,y \in G$. $\endgroup$ – user53970 Nov 5 '13 at 1:50
  • $\begingroup$ @user53970 , This is because if $h \in <x> \cap <y>$ then $h^n \in <x> \cap <y>$ and since $x,y$ have the same order we see either $<x> = <y>$ or $<x> \cap <y>= \{e\}$ right? Thus any two subgroups of order 11 has a trivial intersection or they are the same. $\endgroup$ – AvatarOfChronos Nov 5 '13 at 2:22

(CW to push this away from the unanswered list)

The argument is correct; possibly it should be added that subgroups of order $11$ either have intersection $\{e\}$ or are the same, so indeed $G$ must be the disjoint union of $\{e\}$ and pairwise disjoint ten element subsets.

The property is easy to prove: if $H_1$ and $H_2$ are subgroups of order $11$, then $H_1\cap H_2$ has order either $1$ or $11$, being a subgroup of $H_1$.


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