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In my group theory class our teacher gave us this statement but I don't understand exactly why it's true.

Let $G$ be a non-abelian group such that every proper subgroup of $G$ is abelian, we can find $a$, $b\in G$ that satisfy: $$G=\langle a,b \rangle$$

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    $\begingroup$ I always find it odd when a question is interesting enough to admit a +14 answer, but not interesting enough to get many votes itself. $\endgroup$ – user1729 Oct 10 at 14:17
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    $\begingroup$ @user1729 I agree. I think the question was on the hot question list, which might explain the upvotes. $\endgroup$ – Ethan Bolker Oct 15 at 2:52
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Since $G$ is nonabelian you can find two elements $a$ and $b$ that don't commute. They generate some subgroup that's not abelian. If all the proper subgroups are abelian then that one must be the whole group.

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