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Proving that the order of an intersection is a common divisor of the order of both sets

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1

$begingroup$

I am looking for pointers on a group theory problem. My knowledge is limited to basic group theory, and in the context of my coursework, I believe I am expected to use Lagrange's theorem and other properties of cosets.

Let H and K be subgroups of a finite group G. Prove that |H $cap$ K| is a common divisor of |H| and |K|.

By Lagrange's Theorem, I know that if H $cap$ K is a subgroup of both H and K, then |H $cap$ K| divides both |H| and |K|. My work so far:

H $cap$ K is non-empty:

Since H, K $leq$ G, H, K $neq$ $emptyset$, and so H $cap$ K $neq$ $emptyset$. (Am I wrong about this?)

H $cap$ K is closed under products:

Let a, b $in$ H $cap$ K

a, b $in$ H and K $implies$ ab $in$ H and K $implies$ ab $in$ H $cap$ K.

H $cap$ K is closed under inversion:

a $in$ H $cap$ K $implies$ a $in$ H, K $implies$ a$^-1$ $in$ H, K $implies$ a$^-1$ $in$ H $cap$ K.

Therefore H $cap$ K is a subgroup of both H and K, and by Lagrange's theorem, |H $cap$ K| divides both |H| and |K|.

I do not feel this proof is rigorous. I am not even sure I was right to assume H $cap$ K was a group. I think I should have taken an approach more directly involving cosets, but this was not obvious to me at the outset. I would be grateful for any advice.

abstract-algebra group-theory elementary-set-theory

share|cite|improve this question

asked Mar 16 at 23:06

W. HongoW. Hongo

254

$endgroup$

add a comment | 

1

$begingroup$

I am looking for pointers on a group theory problem. My knowledge is limited to basic group theory, and in the context of my coursework, I believe I am expected to use Lagrange's theorem and other properties of cosets.

Let H and K be subgroups of a finite group G. Prove that |H $cap$ K| is a common divisor of |H| and |K|.

By Lagrange's Theorem, I know that if H $cap$ K is a subgroup of both H and K, then |H $cap$ K| divides both |H| and |K|. My work so far:

H $cap$ K is non-empty:

Since H, K $leq$ G, H, K $neq$ $emptyset$, and so H $cap$ K $neq$ $emptyset$. (Am I wrong about this?)

H $cap$ K is closed under products:

Let a, b $in$ H $cap$ K

a, b $in$ H and K $implies$ ab $in$ H and K $implies$ ab $in$ H $cap$ K.

H $cap$ K is closed under inversion:

a $in$ H $cap$ K $implies$ a $in$ H, K $implies$ a$^-1$ $in$ H, K $implies$ a$^-1$ $in$ H $cap$ K.

Therefore H $cap$ K is a subgroup of both H and K, and by Lagrange's theorem, |H $cap$ K| divides both |H| and |K|.

I do not feel this proof is rigorous. I am not even sure I was right to assume H $cap$ K was a group. I think I should have taken an approach more directly involving cosets, but this was not obvious to me at the outset. I would be grateful for any advice.

abstract-algebra group-theory elementary-set-theory

share|cite|improve this question

asked Mar 16 at 23:06

W. HongoW. Hongo

254

$endgroup$

add a comment | 

$begingroup$

I am looking for pointers on a group theory problem. My knowledge is limited to basic group theory, and in the context of my coursework, I believe I am expected to use Lagrange's theorem and other properties of cosets.

Let H and K be subgroups of a finite group G. Prove that |H $cap$ K| is a common divisor of |H| and |K|.

By Lagrange's Theorem, I know that if H $cap$ K is a subgroup of both H and K, then |H $cap$ K| divides both |H| and |K|. My work so far:

H $cap$ K is non-empty:

Since H, K $leq$ G, H, K $neq$ $emptyset$, and so H $cap$ K $neq$ $emptyset$. (Am I wrong about this?)

H $cap$ K is closed under products:

Let a, b $in$ H $cap$ K

a, b $in$ H and K $implies$ ab $in$ H and K $implies$ ab $in$ H $cap$ K.

H $cap$ K is closed under inversion:

a $in$ H $cap$ K $implies$ a $in$ H, K $implies$ a$^-1$ $in$ H, K $implies$ a$^-1$ $in$ H $cap$ K.

Therefore H $cap$ K is a subgroup of both H and K, and by Lagrange's theorem, |H $cap$ K| divides both |H| and |K|.

I do not feel this proof is rigorous. I am not even sure I was right to assume H $cap$ K was a group. I think I should have taken an approach more directly involving cosets, but this was not obvious to me at the outset. I would be grateful for any advice.

abstract-algebra group-theory elementary-set-theory

share|cite|improve this question

asked Mar 16 at 23:06

W. HongoW. Hongo

254

$endgroup$

share|cite|improve this question

asked Mar 16 at 23:06

W. HongoW. Hongo

254

share|cite|improve this question

asked Mar 16 at 23:06

W. HongoW. Hongo

254

asked Mar 16 at 23:06

W. HongoW. Hongo

254

asked Mar 16 at 23:06

W. HongoW. Hongo

254

1 Answer
1

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oldest

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2

$begingroup$

Yes, $Hcap K$ is a subgroup, it's nonempty, as surely contains the identity element, and Lagrange's theorem applies.

That's it.

share|cite|improve this answer

answered Mar 16 at 23:28

BerciBerci

61.7k23674

$endgroup$

add a comment | 

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2

$begingroup$

Yes, $Hcap K$ is a subgroup, it's nonempty, as surely contains the identity element, and Lagrange's theorem applies.

That's it.

share|cite|improve this answer

answered Mar 16 at 23:28

BerciBerci

61.7k23674

$endgroup$

add a comment | 

2

$begingroup$

Yes, $Hcap K$ is a subgroup, it's nonempty, as surely contains the identity element, and Lagrange's theorem applies.

That's it.

share|cite|improve this answer

answered Mar 16 at 23:28

BerciBerci

61.7k23674

$endgroup$

add a comment | 

$begingroup$

Yes, $Hcap K$ is a subgroup, it's nonempty, as surely contains the identity element, and Lagrange's theorem applies.

That's it.

share|cite|improve this answer

answered Mar 16 at 23:28

BerciBerci

61.7k23674

$endgroup$

share|cite|improve this answer

answered Mar 16 at 23:28

BerciBerci

61.7k23674

answered Mar 16 at 23:28

BerciBerci

61.7k23674

answered Mar 16 at 23:28

BerciBerci

61.7k23674