I didn't think I'd talk about this either up until a few minutes ago.
The Romanian Maths Curriculum is extremely intriguing, mainly because of the fact that we do Analysis (at least Olympiad students do) and Linear and Abstract Algebra during our high school years. From what I've gathered, in College, we go again through these subjects in the first year, but at a more advanced level. The students who attended the Maths Olympiad during high school have a one year head-start over the ones that didn't, if you think about it, and that doesn't apply only for the Romanian Olympians who go to Universities in Romania. Participating at the Olympiad helps greatly in College, as the Olympians have more time to socialize and focus on extra-curricular activities during the first year, because they already know a lot of what the teacher is showing during the first few months. There is, however, a downside; only a few other people will have time to socialize with the Olympians, because in most Universities, there aren't a lot of people who already know a big chuck of the Curriculum.
I stumbled upon this group theory problem today, while going through a popular abstract algebra book. The problem itself, I think, is rather well known, but I discovered something that surprised me about the groups that the problem discusses.
Let \((G,*)\) be a group of order 2n, and \((H,*)\) and \((K,*)\) normal subgroups of order n such that \(H∩K={e}\). Prove that G is abelian.
Solution:
I shall prove that the only group satisfying these conditions is G={e, a, b, ab}, with \(a^2=b^2=e\), and a and b commute.
Since H and K have order n, and their intersection is e, their reunion has 2n-1 elements. Let x be the element that isn't in \(H∪K\). As a fun fact, since x isn't in \(H∪K\), \(x^{-1}\) isn't in \(H∪K\) either, because then x would be in either H or K, contradiction. Thus, \(x=x^{-1}\), so \(x^2=e\). This, however, doesn't really impact our proof. For the rest of the proof, we shall consider H* and K* as H and K minus {e}.
Let \(f_x:H^*⟶K^*\), \(f_x(h)=xh\). Obviously, this function is injective, and, since \(H^*\) and \(K^*\) have the same cardinal, it's bijective. Also, the function is well-defined, because \(xh\) can't be in H, because, if \(xh_1=h_2\), then \(x=h_2*(h_1)^{-1}\), so x would be in H.
Since the function is bijective, we have that \(xh_i=k_i\), for any i between 1 and n-1. Thus, \(x=k_i*(h_i)^{-1}=k_j*(h_j)^{-1}\), for any i and j between 1 and n-1, so \((h_i)^{-1}h_j=(k_i)^{-1}k_j\). The RHS belongs to H, and the LHS belongs to K, and we know that the only element of the intersection between H and K is e, so \(h_i=h_j=h, k_i=k_j=k\) for any i and j. Thus, H={e,h} and K={e,k}, while G is isomorphic with Klein's group: {e, h, k, hk}, and, as a fun fact, G is abelian, as required. The solution that the author of the book offered relied mostly on the fact that x has order 2, and that H and K were normal groups. The problem, as stated in the book, didn't say that H and K were normal subgroups, but rather asked you to prove that in the first part of the problem.
I've got to say, this result is a pretty intriguing property of Klein's group if you think about it. I truly hope that I didn't get anything wrong in the proof. Of course, this result pales in front of theorems such as the classification of simple finite groups, which took thousands of pages to write. However, it is a modest start. Tune in next time, for either another Analytical Geometry problem, another Probability problem, that personality article I kept talking about, or something totally random (again).
Cris.
Nice problem.
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