Mixing
Wreath Product of Two Finitely Generated Groups
$begingroup$
Let $G$ and $H$ be two finitely generated groups, and let $W = G wr H$ be the wreath product of $G$ and $H$. Show that $W$ is finitely generated. In class today, we were showed this and told that it was obvious. However, I do not see how it is obvious. How is this obvious?
abstract-algebra group-theory geometric-group-theory wreath-product
edited Jan 23 at 15:41
Martin Sleziak
44.8k10119273
asked Apr 11 '12 at 0:08
BernardBernard
161
$endgroup$
|
show 4 more comments
$begingroup$
Let $G$ and $H$ be two finitely generated groups, and let $W = G wr H$ be the wreath product of $G$ and $H$. Show that $W$ is finitely generated. In class today, we were showed this and told that it was obvious. However, I do not see how it is obvious. How is this obvious?
abstract-algebra group-theory geometric-group-theory wreath-product
edited Jan 23 at 15:41
Martin Sleziak
44.8k10119273
asked Apr 11 '12 at 0:08
BernardBernard
161
$endgroup$
1
$begingroup$
Take the generators of $G$ together with the generators of $H$. Much harder question is when this is finitely presented...
$endgroup$
– user641
Apr 11 '12 at 0:17
1
$begingroup$
@Alex: is your comment directed at the point that you can either take the direct product or the direct sum in the definition of the wreath product? Of course it must be the latter if you want to have any chance of being finitely generated and then the result is true.
$endgroup$
– t.b.
Apr 11 '12 at 3:04
1
$begingroup$
Well, it is not exactly obvious but it is not hard. I suggest you work through the simplest example: the Lamplighter Group $L = mathbb{Z}/2mathbb{Z} wr mathbb{Z}$. If you manage to prove that the generators of $mathbb{Z}/2mathbb{Z}$ and $mathbb{Z}$ generate $L$ then the general case should be rather easy to do.
$endgroup$
– t.b.
Apr 13 '12 at 1:28
1
$begingroup$
@Bernard: You should specify if you are talking about the "restricted wreath product" (the base group is the restricted direct product of $|H|$ copies of $G$, i.e., the subgroup of $prod_{hin H}G$ with almost all entries trivial), or the "unrestricted wreath product" (the base group is the direct product of $|H|$ copies of $G$).
$endgroup$
– Arturo Magidin
Apr 13 '12 at 2:56
1
$begingroup$
@Arturo: Well, in the comments Bernard says that $H$ is finite, in which case it doesn't matter. But I don't know why he says that, when it isn't in the question and, provided he means the restricted wreath product, it isn't necessary.
$endgroup$
– Tara B
Apr 13 '12 at 8:02
|
show 4 more comments
$begingroup$
Let $G$ and $H$ be two finitely generated groups, and let $W = G wr H$ be the wreath product of $G$ and $H$. Show that $W$ is finitely generated. In class today, we were showed this and told that it was obvious. However, I do not see how it is obvious. How is this obvious?
abstract-algebra group-theory geometric-group-theory wreath-product
edited Jan 23 at 15:41
Martin Sleziak
44.8k10119273
asked Apr 11 '12 at 0:08
BernardBernard
161
$endgroup$
Let $G$ and $H$ be two finitely generated groups, and let $W = G wr H$ be the wreath product of $G$ and $H$. Show that $W$ is finitely generated. In class today, we were showed this and told that it was obvious. However, I do not see how it is obvious. How is this obvious?
abstract-algebra group-theory geometric-group-theory wreath-product
abstract-algebra group-theory geometric-group-theory wreath-product
edited Jan 23 at 15:41
Martin Sleziak
44.8k10119273
asked Apr 11 '12 at 0:08
BernardBernard
161
edited Jan 23 at 15:41
Martin Sleziak
44.8k10119273
asked Apr 11 '12 at 0:08
BernardBernard
161
edited Jan 23 at 15:41
Martin Sleziak
44.8k10119273
edited Jan 23 at 15:41
Martin Sleziak
44.8k10119273
edited Jan 23 at 15:41
Martin Sleziak
44.8k10119273
44.8k10119273
asked Apr 11 '12 at 0:08
BernardBernard
161
asked Apr 11 '12 at 0:08
BernardBernard
161
asked Apr 11 '12 at 0:08
BernardBernard
161
161
1
$begingroup$
Take the generators of $G$ together with the generators of $H$. Much harder question is when this is finitely presented...
$endgroup$
– user641
Apr 11 '12 at 0:17
1
$begingroup$
@Alex: is your comment directed at the point that you can either take the direct product or the direct sum in the definition of the wreath product? Of course it must be the latter if you want to have any chance of being finitely generated and then the result is true.
$endgroup$
– t.b.
Apr 11 '12 at 3:04
1
$begingroup$
Well, it is not exactly obvious but it is not hard. I suggest you work through the simplest example: the Lamplighter Group $L = mathbb{Z}/2mathbb{Z} wr mathbb{Z}$. If you manage to prove that the generators of $mathbb{Z}/2mathbb{Z}$ and $mathbb{Z}$ generate $L$ then the general case should be rather easy to do.
$endgroup$
– t.b.
Apr 13 '12 at 1:28
1
$begingroup$
@Bernard: You should specify if you are talking about the "restricted wreath product" (the base group is the restricted direct product of $|H|$ copies of $G$, i.e., the subgroup of $prod_{hin H}G$ with almost all entries trivial), or the "unrestricted wreath product" (the base group is the direct product of $|H|$ copies of $G$).
$endgroup$
– Arturo Magidin
Apr 13 '12 at 2:56
1
$begingroup$
@Arturo: Well, in the comments Bernard says that $H$ is finite, in which case it doesn't matter. But I don't know why he says that, when it isn't in the question and, provided he means the restricted wreath product, it isn't necessary.
$endgroup$
– Tara B
Apr 13 '12 at 8:02
|
show 4 more comments
1
$begingroup$
Take the generators of $G$ together with the generators of $H$. Much harder question is when this is finitely presented...
$endgroup$
– user641
Apr 11 '12 at 0:17
1
$begingroup$
@Alex: is your comment directed at the point that you can either take the direct product or the direct sum in the definition of the wreath product? Of course it must be the latter if you want to have any chance of being finitely generated and then the result is true.
$endgroup$
– t.b.
Apr 11 '12 at 3:04
1
$begingroup$
Well, it is not exactly obvious but it is not hard. I suggest you work through the simplest example: the Lamplighter Group $L = mathbb{Z}/2mathbb{Z} wr mathbb{Z}$. If you manage to prove that the generators of $mathbb{Z}/2mathbb{Z}$ and $mathbb{Z}$ generate $L$ then the general case should be rather easy to do.
$endgroup$
– t.b.
Apr 13 '12 at 1:28
1
$begingroup$
@Bernard: You should specify if you are talking about the "restricted wreath product" (the base group is the restricted direct product of $|H|$ copies of $G$, i.e., the subgroup of $prod_{hin H}G$ with almost all entries trivial), or the "unrestricted wreath product" (the base group is the direct product of $|H|$ copies of $G$).
$endgroup$
– Arturo Magidin
Apr 13 '12 at 2:56
1
$begingroup$
@Arturo: Well, in the comments Bernard says that $H$ is finite, in which case it doesn't matter. But I don't know why he says that, when it isn't in the question and, provided he means the restricted wreath product, it isn't necessary.
$endgroup$
– Tara B
Apr 13 '12 at 8:02
1
1
$begingroup$
Take the generators of $G$ together with the generators of $H$. Much harder question is when this is finitely presented...
$endgroup$
– user641
Apr 11 '12 at 0:17
$begingroup$
Take the generators of $G$ together with the generators of $H$. Much harder question is when this is finitely presented...
$endgroup$
– user641
Apr 11 '12 at 0:17
1
1
$begingroup$
@Alex: is your comment directed at the point that you can either take the direct product or the direct sum in the definition of the wreath product? Of course it must be the latter if you want to have any chance of being finitely generated and then the result is true.
$endgroup$
– t.b.
Apr 11 '12 at 3:04
$begingroup$
@Alex: is your comment directed at the point that you can either take the direct product or the direct sum in the definition of the wreath product? Of course it must be the latter if you want to have any chance of being finitely generated and then the result is true.
$endgroup$
– t.b.
Apr 11 '12 at 3:04
1
1
$begingroup$
Well, it is not exactly obvious but it is not hard. I suggest you work through the simplest example: the Lamplighter Group $L = mathbb{Z}/2mathbb{Z} wr mathbb{Z}$. If you manage to prove that the generators of $mathbb{Z}/2mathbb{Z}$ and $mathbb{Z}$ generate $L$ then the general case should be rather easy to do.
$endgroup$
– t.b.
Apr 13 '12 at 1:28
$begingroup$
Well, it is not exactly obvious but it is not hard. I suggest you work through the simplest example: the Lamplighter Group $L = mathbb{Z}/2mathbb{Z} wr mathbb{Z}$. If you manage to prove that the generators of $mathbb{Z}/2mathbb{Z}$ and $mathbb{Z}$ generate $L$ then the general case should be rather easy to do.
$endgroup$
– t.b.
Apr 13 '12 at 1:28
1
1
$begingroup$
@Bernard: You should specify if you are talking about the "restricted wreath product" (the base group is the restricted direct product of $|H|$ copies of $G$, i.e., the subgroup of $prod_{hin H}G$ with almost all entries trivial), or the "unrestricted wreath product" (the base group is the direct product of $|H|$ copies of $G$).
$endgroup$
– Arturo Magidin
Apr 13 '12 at 2:56
$begingroup$
@Bernard: You should specify if you are talking about the "restricted wreath product" (the base group is the restricted direct product of $|H|$ copies of $G$, i.e., the subgroup of $prod_{hin H}G$ with almost all entries trivial), or the "unrestricted wreath product" (the base group is the direct product of $|H|$ copies of $G$).
$endgroup$
– Arturo Magidin
Apr 13 '12 at 2:56
1
1
$begingroup$
@Arturo: Well, in the comments Bernard says that $H$ is finite, in which case it doesn't matter. But I don't know why he says that, when it isn't in the question and, provided he means the restricted wreath product, it isn't necessary.
$endgroup$
– Tara B
Apr 13 '12 at 8:02
$begingroup$
@Arturo: Well, in the comments Bernard says that $H$ is finite, in which case it doesn't matter. But I don't know why he says that, when it isn't in the question and, provided he means the restricted wreath product, it isn't necessary.
$endgroup$
– Tara B
Apr 13 '12 at 8:02
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Consider the set of elements of the form $(a,e_H)$ where
$$begin{cases}
a_h in langle G rangle & h = e_H \
a_h = e_G & text{otherwise}
end{cases}$$
together with the set of elements of the form ${e_G}^H times langle H rangle$. More concisely,
$$langle G wr H rangle = left(left({e_G}^{H-{e_H}} times langle G rangle^{{e_H}}right) times {e_H}right) cup left({e_G}^H times langle H rangleright)$$
Then, assuming $e_G notin langle G rangle lor e_H notin langle H rangle$,
begin{align}
|langle G wr H rangle|
&= |left(left({e_G}^{H-{e_H}} times langle G rangle^{{e_H}}right) times {e_H}right) cup left({e_G}^H times langle H rangleright)| \
&= |left({e_G}^{H-{e_H}} times langle G rangle^{{e_H}}right) times {e_H}| + |{e_G}^H times langle H rangle| \
&= |{e_G}^{H-{e_H}}| |langle G rangle^{{e_H}}| |{e_H}| + |{e_G}^H| |langle H rangle| \
&= |{e_G}|^{|H - {e_H}|} |langle G rangle|^{|{e_H}|} 1 + |{e_G}|^{|H|} |langle H rangle| \
&= 1^{|H-{e_H}|} |langle G rangle|^1 + 1^{|H|} |langle H rangle| \
&= 1 |langle G rangle| + 1 |langle H rangle| \
&= |langle G rangle| + |langle H rangle|
end{align}
Since $langle G rangle$ and $langle H rangle$ are finite, $langle G wr H rangle$ is also finite.
answered Jan 21 at 17:12
user76284user76284
1,2231126
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f130245%2fwreath-product-of-two-finitely-generated-groups%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the set of elements of the form $(a,e_H)$ where
$$begin{cases}
a_h in langle G rangle & h = e_H \
a_h = e_G & text{otherwise}
end{cases}$$
together with the set of elements of the form ${e_G}^H times langle H rangle$. More concisely,
$$langle G wr H rangle = left(left({e_G}^{H-{e_H}} times langle G rangle^{{e_H}}right) times {e_H}right) cup left({e_G}^H times langle H rangleright)$$
Then, assuming $e_G notin langle G rangle lor e_H notin langle H rangle$,
begin{align}
|langle G wr H rangle|
&= |left(left({e_G}^{H-{e_H}} times langle G rangle^{{e_H}}right) times {e_H}right) cup left({e_G}^H times langle H rangleright)| \
&= |left({e_G}^{H-{e_H}} times langle G rangle^{{e_H}}right) times {e_H}| + |{e_G}^H times langle H rangle| \
&= |{e_G}^{H-{e_H}}| |langle G rangle^{{e_H}}| |{e_H}| + |{e_G}^H| |langle H rangle| \
&= |{e_G}|^{|H - {e_H}|} |langle G rangle|^{|{e_H}|} 1 + |{e_G}|^{|H|} |langle H rangle| \
&= 1^{|H-{e_H}|} |langle G rangle|^1 + 1^{|H|} |langle H rangle| \
&= 1 |langle G rangle| + 1 |langle H rangle| \
&= |langle G rangle| + |langle H rangle|
end{align}
Since $langle G rangle$ and $langle H rangle$ are finite, $langle G wr H rangle$ is also finite.
answered Jan 21 at 17:12
user76284user76284
1,2231126
$endgroup$
add a comment |
$begingroup$
Consider the set of elements of the form $(a,e_H)$ where
$$begin{cases}
a_h in langle G rangle & h = e_H \
a_h = e_G & text{otherwise}
end{cases}$$
together with the set of elements of the form ${e_G}^H times langle H rangle$. More concisely,
$$langle G wr H rangle = left(left({e_G}^{H-{e_H}} times langle G rangle^{{e_H}}right) times {e_H}right) cup left({e_G}^H times langle H rangleright)$$
Then, assuming $e_G notin langle G rangle lor e_H notin langle H rangle$,
begin{align}
|langle G wr H rangle|
&= |left(left({e_G}^{H-{e_H}} times langle G rangle^{{e_H}}right) times {e_H}right) cup left({e_G}^H times langle H rangleright)| \
&= |left({e_G}^{H-{e_H}} times langle G rangle^{{e_H}}right) times {e_H}| + |{e_G}^H times langle H rangle| \
&= |{e_G}^{H-{e_H}}| |langle G rangle^{{e_H}}| |{e_H}| + |{e_G}^H| |langle H rangle| \
&= |{e_G}|^{|H - {e_H}|} |langle G rangle|^{|{e_H}|} 1 + |{e_G}|^{|H|} |langle H rangle| \
&= 1^{|H-{e_H}|} |langle G rangle|^1 + 1^{|H|} |langle H rangle| \
&= 1 |langle G rangle| + 1 |langle H rangle| \
&= |langle G rangle| + |langle H rangle|
end{align}
Since $langle G rangle$ and $langle H rangle$ are finite, $langle G wr H rangle$ is also finite.
answered Jan 21 at 17:12
user76284user76284
1,2231126
$endgroup$
add a comment |
$begingroup$
Consider the set of elements of the form $(a,e_H)$ where
$$begin{cases}
a_h in langle G rangle & h = e_H \
a_h = e_G & text{otherwise}
end{cases}$$
together with the set of elements of the form ${e_G}^H times langle H rangle$. More concisely,
$$langle G wr H rangle = left(left({e_G}^{H-{e_H}} times langle G rangle^{{e_H}}right) times {e_H}right) cup left({e_G}^H times langle H rangleright)$$
Then, assuming $e_G notin langle G rangle lor e_H notin langle H rangle$,
begin{align}
|langle G wr H rangle|
&= |left(left({e_G}^{H-{e_H}} times langle G rangle^{{e_H}}right) times {e_H}right) cup left({e_G}^H times langle H rangleright)| \
&= |left({e_G}^{H-{e_H}} times langle G rangle^{{e_H}}right) times {e_H}| + |{e_G}^H times langle H rangle| \
&= |{e_G}^{H-{e_H}}| |langle G rangle^{{e_H}}| |{e_H}| + |{e_G}^H| |langle H rangle| \
&= |{e_G}|^{|H - {e_H}|} |langle G rangle|^{|{e_H}|} 1 + |{e_G}|^{|H|} |langle H rangle| \
&= 1^{|H-{e_H}|} |langle G rangle|^1 + 1^{|H|} |langle H rangle| \
&= 1 |langle G rangle| + 1 |langle H rangle| \
&= |langle G rangle| + |langle H rangle|
end{align}
Since $langle G rangle$ and $langle H rangle$ are finite, $langle G wr H rangle$ is also finite.
answered Jan 21 at 17:12
user76284user76284
1,2231126
$endgroup$
Consider the set of elements of the form $(a,e_H)$ where
$$begin{cases}
a_h in langle G rangle & h = e_H \
a_h = e_G & text{otherwise}
end{cases}$$
together with the set of elements of the form ${e_G}^H times langle H rangle$. More concisely,
$$langle G wr H rangle = left(left({e_G}^{H-{e_H}} times langle G rangle^{{e_H}}right) times {e_H}right) cup left({e_G}^H times langle H rangleright)$$
Then, assuming $e_G notin langle G rangle lor e_H notin langle H rangle$,
begin{align}
|langle G wr H rangle|
&= |left(left({e_G}^{H-{e_H}} times langle G rangle^{{e_H}}right) times {e_H}right) cup left({e_G}^H times langle H rangleright)| \
&= |left({e_G}^{H-{e_H}} times langle G rangle^{{e_H}}right) times {e_H}| + |{e_G}^H times langle H rangle| \
&= |{e_G}^{H-{e_H}}| |langle G rangle^{{e_H}}| |{e_H}| + |{e_G}^H| |langle H rangle| \
&= |{e_G}|^{|H - {e_H}|} |langle G rangle|^{|{e_H}|} 1 + |{e_G}|^{|H|} |langle H rangle| \
&= 1^{|H-{e_H}|} |langle G rangle|^1 + 1^{|H|} |langle H rangle| \
&= 1 |langle G rangle| + 1 |langle H rangle| \
&= |langle G rangle| + |langle H rangle|
end{align}
Since $langle G rangle$ and $langle H rangle$ are finite, $langle G wr H rangle$ is also finite.
answered Jan 21 at 17:12
user76284user76284
1,2231126
edited Jan 21 at 17:26
answered Jan 21 at 17:12
user76284user76284
1,2231126
answered Jan 21 at 17:12
user76284user76284
1,2231126
answered Jan 21 at 17:12
user76284user76284
1,2231126
1,2231126
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f130245%2fwreath-product-of-two-finitely-generated-groups%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Take the generators of $G$ together with the generators of $H$. Much harder question is when this is finitely presented...
$endgroup$
– user641
Apr 11 '12 at 0:17
1
$begingroup$
@Alex: is your comment directed at the point that you can either take the direct product or the direct sum in the definition of the wreath product? Of course it must be the latter if you want to have any chance of being finitely generated and then the result is true.
$endgroup$
– t.b.
Apr 11 '12 at 3:04
1
$begingroup$
Well, it is not exactly obvious but it is not hard. I suggest you work through the simplest example: the Lamplighter Group $L = mathbb{Z}/2mathbb{Z} wr mathbb{Z}$. If you manage to prove that the generators of $mathbb{Z}/2mathbb{Z}$ and $mathbb{Z}$ generate $L$ then the general case should be rather easy to do.
$endgroup$
– t.b.
Apr 13 '12 at 1:28
1
$begingroup$
@Bernard: You should specify if you are talking about the "restricted wreath product" (the base group is the restricted direct product of $|H|$ copies of $G$, i.e., the subgroup of $prod_{hin H}G$ with almost all entries trivial), or the "unrestricted wreath product" (the base group is the direct product of $|H|$ copies of $G$).
$endgroup$
– Arturo Magidin
Apr 13 '12 at 2:56
1
$begingroup$
@Arturo: Well, in the comments Bernard says that $H$ is finite, in which case it doesn't matter. But I don't know why he says that, when it isn't in the question and, provided he means the restricted wreath product, it isn't necessary.
$endgroup$
– Tara B
Apr 13 '12 at 8:02