# Proving a group of order $77$ has a subgroup of order $7$ without Sylow theorem.

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.

• you missed $77$ as an order of $x$ – Kushal Bhuyan Dec 16 '15 at 3:36
• 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 $77-1$. – Slade Dec 16 '15 at 3:37
• 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. – user217285 Dec 16 '15 at 3:39
• 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? – Steven Stadnicki Dec 16 '15 at 3:48
• Yes thank you I completely forgot 77. Brain fart! – 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.

DONE.

• Could you explain $o(a) = 11$ case? How do we know $H \neq K$ and that there are two different generators? – user299046 Dec 16 '15 at 3:55
• 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 $o(G)=77$ – Learnmore Dec 16 '15 at 4:05
• But how can I say that $HK$ is a subgroup of G? – user299046 Dec 16 '15 at 4:25
• not a subgroup but atleast a subset;cardinality of a subset cant be greater than the set – Learnmore Dec 16 '15 at 4:41
• 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.) – 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.

• How do you know cardinality of $G - \{e\}$ must be a multiple 10. – user299046 Dec 16 '15 at 5:39
• @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?) – Steven Stadnicki Dec 16 '15 at 6:44
• 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? – user299046 Dec 16 '15 at 6:48
• @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. – Steven Stadnicki Dec 16 '15 at 6:54
• (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.) – Steven Stadnicki Dec 16 '15 at 6:55