Showing $Ha=H$ if and only if $a$ belongs to $H$.$gN = hN$ if and only if $g^-1 h in N$$H$ is a maximal normal subgroup of $G$ if and only if $G/H$ is simple.Show that there is just one subgroup $H subset S_4$ such that $[S_4:H] = 2$If a finite set $G$ is closed under an associative product and both cancellation laws hold, then it is a groupShowing that the intersection of 2 subgroups is a subgroupShow that $G/N$ acts faithfully on $S$ if and only if $N=kerphi$Prove that it is impossible that every non-identity element of $G$ has an order of $2$.Explanations on the proof of Theorem 2.5 in Hungerford's algebraShow $H$ is the only subgroup in $G$ of index 2, when $|G| not = 4$ and $[G:H]=2$A group $G$ is abelian if and only if a certain subset of the direct product is a subgroupNon-identity element in a group has infinite order
Drawing ramified coverings with tikz
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Showing $Ha=H$ if and only if $a$ belongs to $H$.
$gN = hN$ if and only if $g^-1 h in N$$H$ is a maximal normal subgroup of $G$ if and only if $G/H$ is simple.Show that there is just one subgroup $H subset S_4$ such that $[S_4:H] = 2$If a finite set $G$ is closed under an associative product and both cancellation laws hold, then it is a groupShowing that the intersection of 2 subgroups is a subgroupShow that $G/N$ acts faithfully on $S$ if and only if $N=kerphi$Prove that it is impossible that every non-identity element of $G$ has an order of $2$.Explanations on the proof of Theorem 2.5 in Hungerford's algebraShow $H$ is the only subgroup in $G$ of index 2, when $|G| not = 4$ and $[G:H]=2$A group $G$ is abelian if and only if a certain subset of the direct product is a subgroupNon-identity element in a group has infinite order
$begingroup$
Let be $H$ a subgroup. Show $Ha=H$ if and only if $a$ belongs to $H$.
Here what I have understand of the proof: $a$ must belong to $Ha$ because $ea=a$ where $e$ is the identity in $H$. So the only way $H=Ha$ is that a belongs to $H$.
Is that correct?
Now suppose that a belongs to H. for every element h in H, $h=(ha^-1)a$. This implies that h belongs to Ha so also a belongs to Ha. This implies H=Ha.
group-theory
asked Mar 16 at 8:12
homunculushomunculus
325
$endgroup$
add a comment |
$begingroup$
Let be $H$ a subgroup. Show $Ha=H$ if and only if $a$ belongs to $H$.
Here what I have understand of the proof: $a$ must belong to $Ha$ because $ea=a$ where $e$ is the identity in $H$. So the only way $H=Ha$ is that a belongs to $H$.
Is that correct?
Now suppose that a belongs to H. for every element h in H, $h=(ha^-1)a$. This implies that h belongs to Ha so also a belongs to Ha. This implies H=Ha.
group-theory
asked Mar 16 at 8:12
homunculushomunculus
325
$endgroup$
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Shaun
Mar 16 at 8:33
$begingroup$
not enough reputation i have
$endgroup$
– homunculus
Mar 16 at 8:34
$begingroup$
Try again, please. It was probably the 15 minute wait timer that stopped you, not lack of reputation.
$endgroup$
– Shaun
Mar 16 at 8:54
$begingroup$
no it is lack of Rep only 15
$endgroup$
– homunculus
Mar 16 at 8:55
add a comment |
$begingroup$
Let be $H$ a subgroup. Show $Ha=H$ if and only if $a$ belongs to $H$.
Here what I have understand of the proof: $a$ must belong to $Ha$ because $ea=a$ where $e$ is the identity in $H$. So the only way $H=Ha$ is that a belongs to $H$.
Is that correct?
Now suppose that a belongs to H. for every element h in H, $h=(ha^-1)a$. This implies that h belongs to Ha so also a belongs to Ha. This implies H=Ha.
group-theory
asked Mar 16 at 8:12
homunculushomunculus
325
$endgroup$
Let be $H$ a subgroup. Show $Ha=H$ if and only if $a$ belongs to $H$.
Here what I have understand of the proof: $a$ must belong to $Ha$ because $ea=a$ where $e$ is the identity in $H$. So the only way $H=Ha$ is that a belongs to $H$.
Is that correct?
Now suppose that a belongs to H. for every element h in H, $h=(ha^-1)a$. This implies that h belongs to Ha so also a belongs to Ha. This implies H=Ha.
group-theory
group-theory
asked Mar 16 at 8:12
homunculushomunculus
325
asked Mar 16 at 8:12
homunculushomunculus
325
edited Mar 16 at 8:29
homunculus
asked Mar 16 at 8:12
homunculushomunculus
325
asked Mar 16 at 8:12
homunculushomunculus
325
asked Mar 16 at 8:12
homunculushomunculus
325
325
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Shaun
Mar 16 at 8:33
$begingroup$
not enough reputation i have
$endgroup$
– homunculus
Mar 16 at 8:34
$begingroup$
Try again, please. It was probably the 15 minute wait timer that stopped you, not lack of reputation.
$endgroup$
– Shaun
Mar 16 at 8:54
$begingroup$
no it is lack of Rep only 15
$endgroup$
– homunculus
Mar 16 at 8:55
add a comment |
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Shaun
Mar 16 at 8:33
$begingroup$
not enough reputation i have
$endgroup$
– homunculus
Mar 16 at 8:34
$begingroup$
Try again, please. It was probably the 15 minute wait timer that stopped you, not lack of reputation.
$endgroup$
– Shaun
Mar 16 at 8:54
$begingroup$
no it is lack of Rep only 15
$endgroup$
– homunculus
Mar 16 at 8:55
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Shaun
Mar 16 at 8:33
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Shaun
Mar 16 at 8:33
$begingroup$
not enough reputation i have
$endgroup$
– homunculus
Mar 16 at 8:34
$begingroup$
not enough reputation i have
$endgroup$
– homunculus
Mar 16 at 8:34
$begingroup$
Try again, please. It was probably the 15 minute wait timer that stopped you, not lack of reputation.
$endgroup$
– Shaun
Mar 16 at 8:54
$begingroup$
Try again, please. It was probably the 15 minute wait timer that stopped you, not lack of reputation.
$endgroup$
– Shaun
Mar 16 at 8:54
$begingroup$
no it is lack of Rep only 15
$endgroup$
– homunculus
Mar 16 at 8:55
$begingroup$
no it is lack of Rep only 15
$endgroup$
– homunculus
Mar 16 at 8:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $ain H$, then every element of $H$ can be written in the form $(ha^-1)ain Ha$ and due to closure, $Ha=H$. Contrarily, if $Ha=H$ then $a=eain Ha=H$.
answered Mar 16 at 8:28
TheSimpliFireTheSimpliFire
12.9k62462
$endgroup$
$begingroup$
$ha^-1$ always belongs to H because by definition of group, the composition of two elements of a group h and a belongs always to the same group H right?
$endgroup$
– homunculus
Mar 16 at 8:46
$begingroup$
Correct, in general for a group $(G,circ)$ and elements $x,yin G$, $xcirc yin G$ to satisfy the closure axiom. Furthermore, $x^-1$ exists to satisfy the inverse axiom.
$endgroup$
– TheSimpliFire
Mar 16 at 8:48
$begingroup$
ok thanks very much
$endgroup$
– homunculus
Mar 16 at 8:50
add a comment |
$begingroup$
Let $ain G$. Then
beginalign
Ha=H &iff Ha=He \
&iff ae^-1in H \
&iff ain H.
endalign
This is due to the more general theorem that $Hx=Hy$ if and only if $xy^-1in H$.
answered Mar 16 at 8:22
ShaunShaun
9,759113684
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $ain H$, then every element of $H$ can be written in the form $(ha^-1)ain Ha$ and due to closure, $Ha=H$. Contrarily, if $Ha=H$ then $a=eain Ha=H$.
answered Mar 16 at 8:28
TheSimpliFireTheSimpliFire
12.9k62462
$endgroup$
$begingroup$
$ha^-1$ always belongs to H because by definition of group, the composition of two elements of a group h and a belongs always to the same group H right?
$endgroup$
– homunculus
Mar 16 at 8:46
$begingroup$
Correct, in general for a group $(G,circ)$ and elements $x,yin G$, $xcirc yin G$ to satisfy the closure axiom. Furthermore, $x^-1$ exists to satisfy the inverse axiom.
$endgroup$
– TheSimpliFire
Mar 16 at 8:48
$begingroup$
ok thanks very much
$endgroup$
– homunculus
Mar 16 at 8:50
add a comment |
$begingroup$
If $ain H$, then every element of $H$ can be written in the form $(ha^-1)ain Ha$ and due to closure, $Ha=H$. Contrarily, if $Ha=H$ then $a=eain Ha=H$.
answered Mar 16 at 8:28
TheSimpliFireTheSimpliFire
12.9k62462
$endgroup$
$begingroup$
$ha^-1$ always belongs to H because by definition of group, the composition of two elements of a group h and a belongs always to the same group H right?
$endgroup$
– homunculus
Mar 16 at 8:46
$begingroup$
Correct, in general for a group $(G,circ)$ and elements $x,yin G$, $xcirc yin G$ to satisfy the closure axiom. Furthermore, $x^-1$ exists to satisfy the inverse axiom.
$endgroup$
– TheSimpliFire
Mar 16 at 8:48
$begingroup$
ok thanks very much
$endgroup$
– homunculus
Mar 16 at 8:50
add a comment |
$begingroup$
If $ain H$, then every element of $H$ can be written in the form $(ha^-1)ain Ha$ and due to closure, $Ha=H$. Contrarily, if $Ha=H$ then $a=eain Ha=H$.
answered Mar 16 at 8:28
TheSimpliFireTheSimpliFire
12.9k62462
$endgroup$
If $ain H$, then every element of $H$ can be written in the form $(ha^-1)ain Ha$ and due to closure, $Ha=H$. Contrarily, if $Ha=H$ then $a=eain Ha=H$.
answered Mar 16 at 8:28
TheSimpliFireTheSimpliFire
12.9k62462
answered Mar 16 at 8:28
TheSimpliFireTheSimpliFire
12.9k62462
answered Mar 16 at 8:28
TheSimpliFireTheSimpliFire
12.9k62462
answered Mar 16 at 8:28
TheSimpliFireTheSimpliFire
12.9k62462
12.9k62462
$begingroup$
$ha^-1$ always belongs to H because by definition of group, the composition of two elements of a group h and a belongs always to the same group H right?
$endgroup$
– homunculus
Mar 16 at 8:46
$begingroup$
Correct, in general for a group $(G,circ)$ and elements $x,yin G$, $xcirc yin G$ to satisfy the closure axiom. Furthermore, $x^-1$ exists to satisfy the inverse axiom.
$endgroup$
– TheSimpliFire
Mar 16 at 8:48
$begingroup$
ok thanks very much
$endgroup$
– homunculus
Mar 16 at 8:50
add a comment |
$begingroup$
$ha^-1$ always belongs to H because by definition of group, the composition of two elements of a group h and a belongs always to the same group H right?
$endgroup$
– homunculus
Mar 16 at 8:46
$begingroup$
Correct, in general for a group $(G,circ)$ and elements $x,yin G$, $xcirc yin G$ to satisfy the closure axiom. Furthermore, $x^-1$ exists to satisfy the inverse axiom.
$endgroup$
– TheSimpliFire
Mar 16 at 8:48
$begingroup$
ok thanks very much
$endgroup$
– homunculus
Mar 16 at 8:50
$begingroup$
$ha^-1$ always belongs to H because by definition of group, the composition of two elements of a group h and a belongs always to the same group H right?
$endgroup$
– homunculus
Mar 16 at 8:46
$begingroup$
$ha^-1$ always belongs to H because by definition of group, the composition of two elements of a group h and a belongs always to the same group H right?
$endgroup$
– homunculus
Mar 16 at 8:46
$begingroup$
Correct, in general for a group $(G,circ)$ and elements $x,yin G$, $xcirc yin G$ to satisfy the closure axiom. Furthermore, $x^-1$ exists to satisfy the inverse axiom.
$endgroup$
– TheSimpliFire
Mar 16 at 8:48
$begingroup$
Correct, in general for a group $(G,circ)$ and elements $x,yin G$, $xcirc yin G$ to satisfy the closure axiom. Furthermore, $x^-1$ exists to satisfy the inverse axiom.
$endgroup$
– TheSimpliFire
Mar 16 at 8:48
$begingroup$
ok thanks very much
$endgroup$
– homunculus
Mar 16 at 8:50
$begingroup$
ok thanks very much
$endgroup$
– homunculus
Mar 16 at 8:50
add a comment |
$begingroup$
Let $ain G$. Then
beginalign
Ha=H &iff Ha=He \
&iff ae^-1in H \
&iff ain H.
endalign
This is due to the more general theorem that $Hx=Hy$ if and only if $xy^-1in H$.
answered Mar 16 at 8:22
ShaunShaun
9,759113684
$endgroup$
add a comment |
$begingroup$
Let $ain G$. Then
beginalign
Ha=H &iff Ha=He \
&iff ae^-1in H \
&iff ain H.
endalign
This is due to the more general theorem that $Hx=Hy$ if and only if $xy^-1in H$.
answered Mar 16 at 8:22
ShaunShaun
9,759113684
$endgroup$
add a comment |
$begingroup$
Let $ain G$. Then
beginalign
Ha=H &iff Ha=He \
&iff ae^-1in H \
&iff ain H.
endalign
This is due to the more general theorem that $Hx=Hy$ if and only if $xy^-1in H$.
answered Mar 16 at 8:22
ShaunShaun
9,759113684
$endgroup$
Let $ain G$. Then
beginalign
Ha=H &iff Ha=He \
&iff ae^-1in H \
&iff ain H.
endalign
This is due to the more general theorem that $Hx=Hy$ if and only if $xy^-1in H$.
answered Mar 16 at 8:22
ShaunShaun
9,759113684
edited Mar 16 at 8:32
answered Mar 16 at 8:22
ShaunShaun
9,759113684
answered Mar 16 at 8:22
ShaunShaun
9,759113684
answered Mar 16 at 8:22
ShaunShaun
9,759113684
9,759113684
add a comment |
add a comment |
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$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Shaun
Mar 16 at 8:33
$begingroup$
not enough reputation i have
$endgroup$
– homunculus
Mar 16 at 8:34
$begingroup$
Try again, please. It was probably the 15 minute wait timer that stopped you, not lack of reputation.
$endgroup$
– Shaun
Mar 16 at 8:54
$begingroup$
no it is lack of Rep only 15
$endgroup$
– homunculus
Mar 16 at 8:55