Order of a subgroup
The answer is 10.
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2 solutions
Why subgroup of order 4 is abelian ?????
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The order of an element is 1,2, or 4 by Lagrange's theorem . The three nontrivial elements have order 2 or 4. If 4, then the group is cyclic and hence abelian. So every nontrivial element has order 2. Take two of them, a and b . Then a b can't be 1 (because a would equal b ) or a (because b would equal 1 ) or b (because a would equal 1 ) so it must be the fourth element. The same argument shows that b a has to be the fourth element. So a b = b a . But this is true for any two nontrivial elements, so all the elements commute with each other.
More generally, any group of order p 2 is abelian. This follows from two lemmas:
(1) The center Z ( G ) of a p -group is nontrivial.
(2) If G / Z ( G ) is cyclic, then G is abelian.
I am new to group theory; here's what I thought For a group G to be abelian, a b = b a is a necessity for all a and b ∈ G or b = a − 1 b a ; for each a if there is a b such that a b = b a then take either of a or b in the set H It follows from the arguement that ∣ H ∣ ≤ 1 0 ,however that hardly proves the answer to be 10
The only possible orders are (proper) divisors of 2 0 , which are 1 , 2 , 4 , 5 , 1 0 . All groups of order 1 , 2 , 4 , 5 are abelian. 1 is trivial, and any group of prime order must be cyclic, generated by any nontrivial element. The fact that groups of order 4 are abelian is slightly more complicated, but essentially follows from writing down the possible multiplication tables.
That leaves 1 0 . For an example of this in action, let G = D 1 0 , the group of rotations and reflections of a regular 1 0 -gon, and let H = D 5 , the group of rotations and reflections of a regular pentagon. Then H is a subgroup of G (consider the rotations/reflections which fix an inscribed pentagon), and ∣ H ∣ = 1 0 , ∣ G ∣ = 2 0 , and H is nonabelian.