Mixing
Index of subgroup is 2 then for any $g$, $g^2$ belongs to subgroup
$begingroup$
If index of a subgroup $H$ is 2, then $g^2 in H$ for every $g$ in G.
Proof:
Since index is 2, there are only two distinct cosets. Now if $g in H$ then it trivially holds because $H$ is a subgroup. Let $H$ and $gH$ be cosets where $g notin H$ therefore given any other coset which is of the latter form then it is equal to $gH$.
Hence,
$gH$=$g^{-1}H$
$g^2H=H$ $implies g^2 in H$
$blacksquare$
abstract-algebra normal-subgroups
edited Jan 6 at 4:40
J. W. Tanner
51010
asked May 3 '17 at 3:14
TutankhamunTutankhamun
129119
$endgroup$
add a comment |
$begingroup$
If index of a subgroup $H$ is 2, then $g^2 in H$ for every $g$ in G.
Proof:
Since index is 2, there are only two distinct cosets. Now if $g in H$ then it trivially holds because $H$ is a subgroup. Let $H$ and $gH$ be cosets where $g notin H$ therefore given any other coset which is of the latter form then it is equal to $gH$.
Hence,
$gH$=$g^{-1}H$
$g^2H=H$ $implies g^2 in H$
$blacksquare$
abstract-algebra normal-subgroups
edited Jan 6 at 4:40
J. W. Tanner
51010
asked May 3 '17 at 3:14
TutankhamunTutankhamun
129119
$endgroup$
$begingroup$
What is your question?
$endgroup$
– Jesko Hüttenhain
May 3 '17 at 6:56
$begingroup$
Is there any fallacy in the proof??
$endgroup$
– Tutankhamun
May 3 '17 at 10:41
$begingroup$
Your proof is correct.
$endgroup$
– Paramanand Singh
May 6 '17 at 2:56
add a comment |
$begingroup$
If index of a subgroup $H$ is 2, then $g^2 in H$ for every $g$ in G.
Proof:
Since index is 2, there are only two distinct cosets. Now if $g in H$ then it trivially holds because $H$ is a subgroup. Let $H$ and $gH$ be cosets where $g notin H$ therefore given any other coset which is of the latter form then it is equal to $gH$.
Hence,
$gH$=$g^{-1}H$
$g^2H=H$ $implies g^2 in H$
$blacksquare$
abstract-algebra normal-subgroups
edited Jan 6 at 4:40
J. W. Tanner
51010
asked May 3 '17 at 3:14
TutankhamunTutankhamun
129119
$endgroup$
If index of a subgroup $H$ is 2, then $g^2 in H$ for every $g$ in G.
Proof:
Since index is 2, there are only two distinct cosets. Now if $g in H$ then it trivially holds because $H$ is a subgroup. Let $H$ and $gH$ be cosets where $g notin H$ therefore given any other coset which is of the latter form then it is equal to $gH$.
Hence,
$gH$=$g^{-1}H$
$g^2H=H$ $implies g^2 in H$
$blacksquare$
abstract-algebra normal-subgroups
abstract-algebra normal-subgroups
edited Jan 6 at 4:40
J. W. Tanner
51010
asked May 3 '17 at 3:14
TutankhamunTutankhamun
129119
edited Jan 6 at 4:40
J. W. Tanner
51010
asked May 3 '17 at 3:14
TutankhamunTutankhamun
129119
edited Jan 6 at 4:40
J. W. Tanner
51010
edited Jan 6 at 4:40
J. W. Tanner
51010
edited Jan 6 at 4:40
J. W. Tanner
51010
51010
asked May 3 '17 at 3:14
TutankhamunTutankhamun
129119
asked May 3 '17 at 3:14
TutankhamunTutankhamun
129119
asked May 3 '17 at 3:14
TutankhamunTutankhamun
129119
129119
$begingroup$
What is your question?
$endgroup$
– Jesko Hüttenhain
May 3 '17 at 6:56
$begingroup$
Is there any fallacy in the proof??
$endgroup$
– Tutankhamun
May 3 '17 at 10:41
$begingroup$
Your proof is correct.
$endgroup$
– Paramanand Singh
May 6 '17 at 2:56
add a comment |
$begingroup$
What is your question?
$endgroup$
– Jesko Hüttenhain
May 3 '17 at 6:56
$begingroup$
Is there any fallacy in the proof??
$endgroup$
– Tutankhamun
May 3 '17 at 10:41
$begingroup$
Your proof is correct.
$endgroup$
– Paramanand Singh
May 6 '17 at 2:56
$begingroup$
What is your question?
$endgroup$
– Jesko Hüttenhain
May 3 '17 at 6:56
$begingroup$
What is your question?
$endgroup$
– Jesko Hüttenhain
May 3 '17 at 6:56
$begingroup$
Is there any fallacy in the proof??
$endgroup$
– Tutankhamun
May 3 '17 at 10:41
$begingroup$
Is there any fallacy in the proof??
$endgroup$
– Tutankhamun
May 3 '17 at 10:41
$begingroup$
Your proof is correct.
$endgroup$
– Paramanand Singh
May 6 '17 at 2:56
$begingroup$
Your proof is correct.
$endgroup$
– Paramanand Singh
May 6 '17 at 2:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Typically one shows that if $|G:H|=2$ then $H$ is normal in $G$. This follows directly since for $gin G$ and $gnotin H$ we have $G = H cup gH = H cup Hg$ which implies $gH = Hg$.
Therefore $G/H cong mathbb{Z}_2$ since there is only one group of order 2. Hence $g^2H = H$ for any $g$ which implies $g^2 in H$.
answered May 4 '17 at 20:47
JJRJJR
1,14828
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
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votes
$begingroup$
Typically one shows that if $|G:H|=2$ then $H$ is normal in $G$. This follows directly since for $gin G$ and $gnotin H$ we have $G = H cup gH = H cup Hg$ which implies $gH = Hg$.
Therefore $G/H cong mathbb{Z}_2$ since there is only one group of order 2. Hence $g^2H = H$ for any $g$ which implies $g^2 in H$.
answered May 4 '17 at 20:47
JJRJJR
1,14828
$endgroup$
add a comment |
$begingroup$
Typically one shows that if $|G:H|=2$ then $H$ is normal in $G$. This follows directly since for $gin G$ and $gnotin H$ we have $G = H cup gH = H cup Hg$ which implies $gH = Hg$.
Therefore $G/H cong mathbb{Z}_2$ since there is only one group of order 2. Hence $g^2H = H$ for any $g$ which implies $g^2 in H$.
answered May 4 '17 at 20:47
JJRJJR
1,14828
$endgroup$
add a comment |
$begingroup$
Typically one shows that if $|G:H|=2$ then $H$ is normal in $G$. This follows directly since for $gin G$ and $gnotin H$ we have $G = H cup gH = H cup Hg$ which implies $gH = Hg$.
Therefore $G/H cong mathbb{Z}_2$ since there is only one group of order 2. Hence $g^2H = H$ for any $g$ which implies $g^2 in H$.
answered May 4 '17 at 20:47
JJRJJR
1,14828
$endgroup$
Typically one shows that if $|G:H|=2$ then $H$ is normal in $G$. This follows directly since for $gin G$ and $gnotin H$ we have $G = H cup gH = H cup Hg$ which implies $gH = Hg$.
Therefore $G/H cong mathbb{Z}_2$ since there is only one group of order 2. Hence $g^2H = H$ for any $g$ which implies $g^2 in H$.
answered May 4 '17 at 20:47
JJRJJR
1,14828
answered May 4 '17 at 20:47
JJRJJR
1,14828
answered May 4 '17 at 20:47
JJRJJR
1,14828
answered May 4 '17 at 20:47
JJRJJR
1,14828
1,14828
add a comment |
add a comment |
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$begingroup$
What is your question?
$endgroup$
– Jesko Hüttenhain
May 3 '17 at 6:56
$begingroup$
Is there any fallacy in the proof??
$endgroup$
– Tutankhamun
May 3 '17 at 10:41
$begingroup$
Your proof is correct.
$endgroup$
– Paramanand Singh
May 6 '17 at 2:56