Intersection of subgroups is a subgroup


by TheoryNoob
Tags: intersection, subgroup, subgroups
TheoryNoob
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#1
Dec2-12, 04:03 AM
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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.
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Fredrik
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#2
Dec2-12, 04:31 AM
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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...
Michael Redei
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#3
Dec2-12, 04:41 AM
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Quote Quote by Fredrik View Post
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?
If for each ##a,b\in G## you can show that ##ab^{-1}\in G##, then you have
  • ##1_G = aa^{-1} \in G##, i.e. ##G## has an identity element.
  • ##b^{-1} = 1_Gb^{-1} \in G##, i.e. ##G## contains an inverse for each element.
  • ##ab = a(b^{-1})^{-1} \in G##, i.e. ##G## is closed under the group operation.

TheoryNoob
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#4
Dec2-12, 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?
Fredrik
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#5
Dec2-12, 04:59 AM
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Quote Quote by Michael Redei View Post
If for each ##a,b\in G## you can show that ##ab^{-1}\in G##, then you have
  • ##1_G = aa^{-1} \in G##, i.e. ##G## has an identity element.
  • ##b^{-1} = 1_Gb^{-1} \in G##, i.e. ##G## contains an inverse for each element.
  • ##ab = a(b^{-1})^{-1} \in G##, i.e. ##G## is closed under the group operation.
Ah, thanks. I suspected something like that, but I didn't remember that theorem and was too lazy to prove it.

Quote Quote by TheoryNoob View Post
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?
Yes, it's fine. But I think you should include something like this: Since ##b\in H## and H is a subgroup, ##b^{-1}\in H##. Since ##b\in K## and K is a subgroup, ##b^{-1}\in K##.
TheoryNoob
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#6
Dec2-12, 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.
Fredrik
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#7
Dec2-12, 05:25 AM
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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##.
TheoryNoob
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#8
Dec2-12, 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.


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