The question is

Show if $G$ has order 77 then $G$ has a subgroup of order 7. Without using Sylow Theorems.

Attempt sketch:

Let $x \in G$. By Lagrange's theorem the order of $x$ is either $1, 7, 11$. Suppose $x \neq e$ Then $x$ has order $7$ or $11$. Now suppose $|x| = 7$. Then $x$ is the generator of a group of order $7$ and we are done. Suppose now that there does not exist an element of order $7$ in G. Then $G$ cannot have order $77$ since $G$ would be cyclic of order $11$ Hence $G$ has an element of order $7$ and thus a subgroup of order $7$.

Please excuse the horrible way this is written.

  • $\begingroup$ you missed $77$ as an order of $x$ $\endgroup$ – Kushal Bhuyan Dec 16 '15 at 3:36
  • $\begingroup$ In addition to the possibility that an element has order $77$, you need to consider the possibility that every element of $G$ has order $1$ or $11$. Hint: $11-1$ is not a factor of $77-1$. $\endgroup$ – Slade Dec 16 '15 at 3:37
  • $\begingroup$ If $x$ has order 77 then $x^{11}$ generates a cyclic group of order 7, so there's no problem there. The real problem is "Then $G$ cannot have order 77 since $G$ would be cyclic of order 11." This really doesn't make any sense since this doesn't actually show that $G$ is cyclic, let alone of order 11. $\endgroup$ – user217285 Dec 16 '15 at 3:39
  • $\begingroup$ Expanding on Slade's comment: suppose that all elements of $G$ have order 11 (since you've already eliminated the cases where some element has order 7 or order 77, by earlier comments). Then can you see how to group these (non-identity) elements into clusters of 10? Can you see how this leads to a contradiction? $\endgroup$ – Steven Stadnicki Dec 16 '15 at 3:48
  • $\begingroup$ Yes thank you I completely forgot 77. Brain fart! $\endgroup$ – user299046 Dec 16 '15 at 3:51

Let $a(\neq e)\in G$ then order of an element divides order of the group ,so possible values of $o(a)=1,7,11,77$.

Now it is not possible that $G$ has all elements of order $11$ because if so then let $o(a)=11,o(b)=11$;take $H=\langle a \rangle ,K=\langle b \rangle $ as $o(H)=o(K)=11 $ and $H\cap K=\{e\}$ then $o(HK)=\dfrac{o(H)o(K)}{o(H\cap K)}=121$.Now $HK\subseteq G$ so do you see a contradiction?

Thus $G$ must have atleast one element of order $7$.

If $o(a)=7\implies $ we have a cyclic group generated by $\langle a \rangle $ of order $7$.

If $o(a)=77$.Then $G$ is cyclic and so $G$ has a unique subgroup corresponding to its each divisor.


  • $\begingroup$ Could you explain $o(a) = 11$ case? How do we know $H \neq K$ and that there are two different generators? $\endgroup$ – user299046 Dec 16 '15 at 3:55
  • $\begingroup$ Definitely; $a$ was chosen to be element of $G$ .Since $a$ has order $11$ it cannot exhaust whole of $G$ since $a^{11}=e$.Hence $G$ has element of some other order say $b$ .If $o(b)=11$ then consider $H,K$ as above we have $H\cap K=\{e\}$.Then $o(HK)=o(H)o(K)=121$ which can't be as $o(G)=77$ $\endgroup$ – Learnmore Dec 16 '15 at 4:05
  • $\begingroup$ But how can I say that $HK$ is a subgroup of G? $\endgroup$ – user299046 Dec 16 '15 at 4:25
  • $\begingroup$ not a subgroup but atleast a subset;cardinality of a subset cant be greater than the set $\endgroup$ – Learnmore Dec 16 '15 at 4:41
  • $\begingroup$ The '11' case here is still pretty sketchily-argued - you've given no reason why $o(HK)$ has to be 121. (Indeed, it's very easy to construct a group with $o(a)=11, o(b)=11$ and $o(ab)=7$, for instance. Such a group can't have order 77, but that's not immediately obvious.) $\endgroup$ – Steven Stadnicki Dec 16 '15 at 5:48

All elements of $G$ must have orders $1$, $7$, $11$, $77$ by Lagrange's Theorem. If $G$ has an element of order $7$, then we're done. If it has an element $a$ of order $77$, then it also has an element of order $7$, namely $a^{11}$.

Therefore we may assume that all elements besides $e$ have order $11$. In that case, $G$ is the union of several subgroups of order $11$, whose pairwise intersections are all $\{e\}$. Thus the cardinality of $G - \{e\}$ must be a multiple of $10$. But this number is actually $76$, a contradiction.

  • $\begingroup$ How do you know cardinality of $G - \{e\}$ must be a multiple 10. $\endgroup$ – user299046 Dec 16 '15 at 5:39
  • $\begingroup$ @user299046 Group all the elements of order 11 in clusters of 10 plus $e$ : if there's an element $a$, then cluster it with $a^2, a^3, \ldots, a^{10}$. (All these clusters are uniquely defined - can you see why?) $\endgroup$ – Steven Stadnicki Dec 16 '15 at 6:44
  • $\begingroup$ Yes each group has a distinct generator. I'm not sure if there must be 7 separate cyclic subgroups of order 11. Is it because all the elements in $G$ need to be accounted for? $\endgroup$ – user299046 Dec 16 '15 at 6:48
  • $\begingroup$ @user299046 The point is that these clusters have to cover all the elements of $G$ (except for the identity). Imagine marking elements - then these clusters mean that you're going to be marking them in groups of ten. But it's impossible to mark 76 elements (all the non-identity elements of $G$) in groups of 10. $\endgroup$ – Steven Stadnicki Dec 16 '15 at 6:54
  • $\begingroup$ (And it's not exactly because each group has a distinct generator - why does, say, $a^4$ have order 11? The fact that 11 is prime is critical here.) $\endgroup$ – Steven Stadnicki Dec 16 '15 at 6:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.