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Quite confused about how to solve this question..As it would have been easier to solve if the group was given to be cyclic, but no such case here

Let $H$ and $K$ be subgroups of a group $G$ of orders $14$ and $21$ respectively. If $H \cap K \neq\{e\}$, here $e$ is the identity of the group $G$, then show that $H \cap K$ is a normal subgroup of $G$.

user1729
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saurabh shukla
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1 Answers1

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This is not true. The issue/trick is that there is no information about the group $G$ so I can pick it to have very few normal subgroups.

For example, we can take $G=S_n$ for some $n\geq 7$ (why?). Then $G$ has very few normal subgroups (see here), and in particular none of order $7$ (why is $7$ relevant?).

For a concrete counter-example, take $H=\langle (1,2,\ldots,7)(8,\ldots,14)\rangle$ and $K=\langle (1,2,\ldots,7)(8,\ldots,21)\rangle$.

user1729
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  • yes i have mentioned this thing in title that no information about G is given.. You may see it – saurabh shukla Nov 15 '19 at 06:48
  • @saurabh do you understand what my answer is saying? – user1729 Nov 15 '19 at 06:52
  • Not actually... Quite confusing, Why are you taking G as symmetric group while no information is given about G..Also we are not disproving it..That we can do by taking counter example..As no information about G is given how we gonna take G in Exam – saurabh shukla Nov 15 '19 at 07:03
  • Because no information is given about $G$ I can pick it to be whatever I want (picking $G$ is really just finding the counter-example). As every group embeds into some symmetric group, the situation in the question must occur in some symmetric group. – user1729 Nov 15 '19 at 07:15
  • Hmm...maybe for a true counter-example you should find the $H$ and $K$. There are easy examples of groups with this property in $S_{21}$, but I'll let you find them (you just need to find some group which has two subgroups with the appropriate property. Combine with my answer to get the counter-example. – user1729 Nov 15 '19 at 07:29
  • @user1729, to complete your counter-example, shouldn't you show that such $S_n$ has actually two subgroups as in the OP? –  Nov 15 '19 at 07:30
  • @Luca yes, see my comment immediately above yours! (Although you don't actually have to show that $S_n$ has two such subgroups, but that some group has! Then embed this into $S_n$.) – user1729 Nov 15 '19 at 07:34
  • @user1729, oh I see! Yes, right. –  Nov 15 '19 at 07:38