This is a question from the free Harvard online abstract algebra lectures. I'm posting my solutions here to get some feedback on them. For a fuller explanation, see this post.

This problem is from assignment 5.

Let $H, K$ be subgroups of a group $G$ of orders 3,5 respectively. Prove that $H\cap K=\{1\}$.

The order of each element of $H$ must divide 3. Since the identity element is the only element with order 1, every other element in $H$ has order 3. Similar reasoning shows every nonidentity element of $K$ has order 5. Since $H$ and $K$ are both subgroups of $G$ they share the same identity element. Therefore, $H\cap K =\{1\}$.

Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem.



Sure, this gets the job done. You can also just observe that the intersection would be a subgroup of both $H$ and $K$. Then using the fact that the order of a subgroup divides the order of the group, this would force the order of the intersection to divide both $3$ and $5$, showing that it is trivial.

Of course, your solution and this one are very much along the same lines; all the exercise is really looking for is to test your understanding of the various divisibility relations that exist in a group. I wrote this up because this avoids having to get down to the level of elements; often cleaner and more structurally-enlightening proofs avoid working at the element level. Here it doesn't make a lick of difference, but it might be good practice for later on, where thinking at a higher structural level can have a much bigger payoff.

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  • $\begingroup$ Thanks. This question is the first time in this course that the intersection of subgroups has been mentioned at all so I wasn't thinking in terms of the intersection being a subgroup. Like you said, it doesn't seem to make much difference in this problem but I can see how looking at these types of problems from that perspective would be less messy in general. $\endgroup$ – jobrien929 Jan 16 '12 at 5:18

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