
#1
Dec212, 04:03 AM

P: 4

1. The problem statement, all variables and given/known data
Suppose H and K are subgroups of G. Prove H intersect K is a subgroup of G. 2. Relevant equations Suppose G is a group and H is a nonempty subset of G. Then H is a subgroup of G iff a,b ∈ H implies ab^1 ∈ H. 3. The attempt at a solution Suppose a and b elements of H intersect K. Since H is a subgroup of G and K is a subgroup G, then ab^1 is an element of H and ab^1 is an element of K, this implies ab^1 is an element of H intersect K, this implies H intersect K is a subgroup of G. 



#2
Dec212, 04:31 AM

Emeritus
Sci Advisor
PF Gold
P: 9,017

You're off to a decent start, but why are you considering ##ab^{1}## instead of ##ab##? What is your definition of "subgroup"? Shouldn't you say something about the identity element and something more specific about inverse elements?
You might want to use shorter sentences, or at least commas. I would start something like this: Let ##a,b\in H\cap K## be arbitrary. Since H is a subgroup and ##a,b\in H##, we have ##ab\in H##. Since K is a subgroup... 



#3
Dec212, 04:41 AM

P: 181





#4
Dec212, 04:46 AM

P: 4

Intersection of subgroups is a subgroup
I'm sorry I should have included this in part 2 of the outline.
I am basing this proof of the following theorem: Suppose G is a group and H is a nonempty subset of G. Then H is a subgroup of G iff a,b ∈ H implies ab^1 ∈ H. We have previously proved this theorem which is why I didn't mention an identity or inverse. Sorry about the long sentence in the proof, I have it written here on paper with symbols and tried to directly type it into words. Based off this theorem is my proof ok? 



#5
Dec212, 04:59 AM

Emeritus
Sci Advisor
PF Gold
P: 9,017





#6
Dec212, 05:06 AM

P: 4

Thanks for the help. The proof seemed to simple so I had to ask, but now that I realize all the real work for the proof is in the previous theorem it makes sense to be so short.




#7
Dec212, 05:25 AM

Emeritus
Sci Advisor
PF Gold
P: 9,017

It's actually just as simple without that theorem.
Without: Let ##a,b\in H\cap K## be arbitrary. Since H is a subgroup, we have ##e\in H##, ##a^{1}\in H## and ##ab\in H##. Since K is a subgroup, we have ##e\in K##, ##a^{1}\in K## and ##ab\in K##. So ##e, a^{1}, ab\in H\cap K##. With: Let ##a,b\in H\cap K## be arbitrary. Since H is a subgroup, ##b^{1}\in H##. So ##a,b^{1}\in H##. Since H is a subgroup, this implies that ##ab^{1}\in H##. Since K is a subgroup, ##b^{1}\in K##. So ##a,b^{1}\in K##. Since K is a subgroup, this implies that ##ab^{1}\in K##. So ##ab^{1}\in H\cap K##. 



#8
Dec212, 05:37 AM

P: 4

Thanks for the help. Your last post has given me the confidence that my answer is correct and helped clarify my understanding of the theorem.



Register to reply 
Related Discussions  
If H and K are subgroups of G, and K is normal, then HK is a subgroup of G.  Calculus & Beyond Homework  1  
Product of two subgroups and intersection with psubgroup  Linear & Abstract Algebra  1  
Is direct product of subgroups a subgroup?  Linear & Abstract Algebra  1  
Prove that the intersection of all subgroups of G of order n is a normal subgroup of  Calculus & Beyond Homework  1  
Intersection of 2 subgroups is a subgroup?  Calculus & Beyond Homework  4 