Prove/Disprove that $left|phi^{-1}({a})right| = left|phi^{-1}({b})right|$ for every $a, b in f(G)$...
$begingroup$
This question already has an answer here:
Number of preimages of a group element under a homomorphism.
1 answer
I encountered this statement of which I don't know if it is true or not:
Let $G, H$ be finite groups and let $phi$ be a homomorphism from G to H. For every $a, b in phi(G)$ it holds true that $left|phi^{-1}({a})right| = left|phi^{-1}({b})right|$.
My question is if this is a valid statement. Furthermore how can one prove/disprove it.
abstract-algebra group-theory
asked Jan 17 at 19:53
qwipoqwipo
265
$endgroup$
marked as duplicate by Dietrich Burde abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 17 at 19:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Number of preimages of a group element under a homomorphism.
1 answer
I encountered this statement of which I don't know if it is true or not:
Let $G, H$ be finite groups and let $phi$ be a homomorphism from G to H. For every $a, b in phi(G)$ it holds true that $left|phi^{-1}({a})right| = left|phi^{-1}({b})right|$.
My question is if this is a valid statement. Furthermore how can one prove/disprove it.
abstract-algebra group-theory
asked Jan 17 at 19:53
qwipoqwipo
265
$endgroup$
marked as duplicate by Dietrich Burde abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 17 at 19:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Number of preimages of a group element under a homomorphism.
1 answer
I encountered this statement of which I don't know if it is true or not:
Let $G, H$ be finite groups and let $phi$ be a homomorphism from G to H. For every $a, b in phi(G)$ it holds true that $left|phi^{-1}({a})right| = left|phi^{-1}({b})right|$.
My question is if this is a valid statement. Furthermore how can one prove/disprove it.
abstract-algebra group-theory
asked Jan 17 at 19:53
qwipoqwipo
265
$endgroup$
This question already has an answer here:
Number of preimages of a group element under a homomorphism.
1 answer
I encountered this statement of which I don't know if it is true or not:
Let $G, H$ be finite groups and let $phi$ be a homomorphism from G to H. For every $a, b in phi(G)$ it holds true that $left|phi^{-1}({a})right| = left|phi^{-1}({b})right|$.
My question is if this is a valid statement. Furthermore how can one prove/disprove it.
This question already has an answer here:
Number of preimages of a group element under a homomorphism.
1 answer
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 17 at 19:53
qwipoqwipo
265
asked Jan 17 at 19:53
qwipoqwipo
265
edited Jan 17 at 19:55
qwipo
asked Jan 17 at 19:53
qwipoqwipo
265
asked Jan 17 at 19:53
qwipoqwipo
265
asked Jan 17 at 19:53
qwipoqwipo
265
265
marked as duplicate by Dietrich Burde abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 17 at 19:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Dietrich Burde abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 17 at 19:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes it's true, and i think the hypotesy of finite groups is unnecessary.
If $phicolon Gto H$ is an homomorphism then $G/ker phicongphi(G)$, that means that $phi^{-1}(g)=gkerphi$.
So $|phi^{-1}(g)|=|kerphi|=|phi^{-1}(0)|:forall gin G$
answered Jan 17 at 20:04
ecrinecrin
3477
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes it's true, and i think the hypotesy of finite groups is unnecessary.
If $phicolon Gto H$ is an homomorphism then $G/ker phicongphi(G)$, that means that $phi^{-1}(g)=gkerphi$.
So $|phi^{-1}(g)|=|kerphi|=|phi^{-1}(0)|:forall gin G$
answered Jan 17 at 20:04
ecrinecrin
3477
$endgroup$
add a comment |
$begingroup$
Yes it's true, and i think the hypotesy of finite groups is unnecessary.
If $phicolon Gto H$ is an homomorphism then $G/ker phicongphi(G)$, that means that $phi^{-1}(g)=gkerphi$.
So $|phi^{-1}(g)|=|kerphi|=|phi^{-1}(0)|:forall gin G$
answered Jan 17 at 20:04
ecrinecrin
3477
$endgroup$
add a comment |
$begingroup$
Yes it's true, and i think the hypotesy of finite groups is unnecessary.
If $phicolon Gto H$ is an homomorphism then $G/ker phicongphi(G)$, that means that $phi^{-1}(g)=gkerphi$.
So $|phi^{-1}(g)|=|kerphi|=|phi^{-1}(0)|:forall gin G$
answered Jan 17 at 20:04
ecrinecrin
3477
$endgroup$
Yes it's true, and i think the hypotesy of finite groups is unnecessary.
If $phicolon Gto H$ is an homomorphism then $G/ker phicongphi(G)$, that means that $phi^{-1}(g)=gkerphi$.
So $|phi^{-1}(g)|=|kerphi|=|phi^{-1}(0)|:forall gin G$
answered Jan 17 at 20:04
ecrinecrin
3477
answered Jan 17 at 20:04
ecrinecrin
3477
answered Jan 17 at 20:04
ecrinecrin
3477
answered Jan 17 at 20:04
ecrinecrin
3477
3477
add a comment |
add a comment |
