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Let $H_{1}$, $H_{2}$ be two distinct subgroups of a finite group $G$, each of order $2$. Let $H$ be the smallest subgroup containing $H_{1}$ and $H_{2}$. Then the order of $H$ is 

  1. Always 2 
  2. Always 4 
  3. Always 8 
  4. None of the above
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From Lagrange's theorem we can say , if H is a subgroup of G, then order of G is divisible by order of H.

Here, G1 and G2 are subgroup of G.  and G is minimum  and containing G1 and G2 .

As, 2 is a prime no. , it cannot be furthur minimized.

So, G can be any multiple of 4. But as the subgroups are distinct ,so it cannot be 2 , but can be 4, 8........

H is smallest subgroup .So, it must be 4

b) is the answer

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The two subgroups are H1={1,x} and H2={1,y}, essentially (with x2=y2=1). They're not really disjoint, as all subgroups contain 1, so I assume we mean that they're disjoint except for that.
So we get in H the elements 1,x,y,xy,yx,xyx,yxy,…There is no natural upper limit (because xy and yx need not have order 2). I think we can go as high as we like using x and y reflections in the plane, and xy becomes a rotation, and we can get lots of finite orders for that...)

If H1 or H2 is normal, then  statement is true, in fact in that case H=H1H2 and |H1H2|=4. In general, the set H1H2 has 4 elements, since |H1H2|=|H1| |H2| / | H1∩H2|=2.2/ 1= 4, and of course H1H2⊆H. So H has at least 4 elements.

Reference : Stackoverflow 

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I think option d. is correct. If H1={e,h1}and H2={e,h2}, what happens if h1 and h2 can't commute? I think the order!
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|H1| = 2 and |H2| = 2 and both are disjoint like no common element between H1 and H2

(A) is false as we | H | >=4 ( Because we want H1 and H2 both set element )

(B) is false if |G| = 10 then we can’t have SubGroup(H) such that |H| = 4 (i.e 4 can’t divide 10 )

(C) is false if |G| = 10 then we can’t have Subgroup(H) such that |H| = 8 (i.e 8 can’t divide 10 )

so left with option D

Correct answer is Option D
Answer:

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