Mixing
Which of these numbers could be the exact number of elements of order $21$ in a group?
$begingroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.46.
Which of the following numbers could be the exact number of elements of order $21$ in a group: $21600, 21602, 21604$?
My Attempt:
Lemma: In a finite group, the number of elements of order $d$ is a multiple of $varphi(d)$.
(This is a Corollary of Theorem 4.4 ibid.)
Since $varphi(21)=12$ and $12mid 21600$ but not the other two candidate numbers, the lemma above implies that $21600$ is the exact number of elements of order $21$ for some (finite) group.
I'm quite sure I got this right$^{dagger}$. I intuited the result, having forgotten the Lemma (well, Corollary) above.
This question is just so I can make note of it, really, but I suppose I could use the post to ask the following:
Is the assumption that the group is finite necessary?
My guess is that it's not. I have a rough idea of using a presentation with some free generators in such a way that it has the required number of order $21$ elements all within some finite subgroup (and no other such elements in the group given by the presentation).
Please help :)
Here $varphi$ is Euler's totient function.
$dagger$ The Dunning-Kruger effect in action . . . It's the right number, yeah, but, as pointed out below, I didn't show that such a group actually exists.
group-theory finite-groups totient-function group-presentation
asked Jan 15 at 14:56
ShaunShaun
9,241113684
$endgroup$
|
show 3 more comments
$begingroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.46.
Which of the following numbers could be the exact number of elements of order $21$ in a group: $21600, 21602, 21604$?
My Attempt:
Lemma: In a finite group, the number of elements of order $d$ is a multiple of $varphi(d)$.
(This is a Corollary of Theorem 4.4 ibid.)
Since $varphi(21)=12$ and $12mid 21600$ but not the other two candidate numbers, the lemma above implies that $21600$ is the exact number of elements of order $21$ for some (finite) group.
I'm quite sure I got this right$^{dagger}$. I intuited the result, having forgotten the Lemma (well, Corollary) above.
This question is just so I can make note of it, really, but I suppose I could use the post to ask the following:
Is the assumption that the group is finite necessary?
My guess is that it's not. I have a rough idea of using a presentation with some free generators in such a way that it has the required number of order $21$ elements all within some finite subgroup (and no other such elements in the group given by the presentation).
Please help :)
Here $varphi$ is Euler's totient function.
$dagger$ The Dunning-Kruger effect in action . . . It's the right number, yeah, but, as pointed out below, I didn't show that such a group actually exists.
group-theory finite-groups totient-function group-presentation
asked Jan 15 at 14:56
ShaunShaun
9,241113684
$endgroup$
1
$begingroup$
I don't know if I'm being naive here, but if you had some infinite group $G$ and you were concerned about the possible precise number of elements of order $d$ that $G$ could have (given that the number of such elements is finite), could you not just consider $H$ to be the smallest subgroup of $G$ containing all elements of order $d$ (or even just the subgroup generated by the elements of order $d$) and then apply your lemma?
$endgroup$
– Adam Higgins
Jan 15 at 15:03
3
$begingroup$
@AdamHiggins $H$ is not necessarily finite: for example, if we take the group with presentation $langle a, b | a^{21}, b^{21}rangle$, then no finite (indeed, no proper) subgroup contains both $a$ and $b$.
$endgroup$
– user3482749
Jan 15 at 15:04
2
$begingroup$
@AdamHiggins $H$ might still be infinite.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 15:04
2
$begingroup$
Thanks all! My bad!
$endgroup$
– Adam Higgins
Jan 15 at 15:05
1
$begingroup$
You also need to show an example to a group such that the number of elements of order $21$ is exactly $21600$. The lemma you cited only shows that the other two possibilities are wrong, it does not give you that the remaining one is surely right.
$endgroup$
– A. Pongrácz
Jan 15 at 15:29
|
show 3 more comments
$begingroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.46.
Which of the following numbers could be the exact number of elements of order $21$ in a group: $21600, 21602, 21604$?
My Attempt:
Lemma: In a finite group, the number of elements of order $d$ is a multiple of $varphi(d)$.
(This is a Corollary of Theorem 4.4 ibid.)
Since $varphi(21)=12$ and $12mid 21600$ but not the other two candidate numbers, the lemma above implies that $21600$ is the exact number of elements of order $21$ for some (finite) group.
I'm quite sure I got this right$^{dagger}$. I intuited the result, having forgotten the Lemma (well, Corollary) above.
This question is just so I can make note of it, really, but I suppose I could use the post to ask the following:
Is the assumption that the group is finite necessary?
My guess is that it's not. I have a rough idea of using a presentation with some free generators in such a way that it has the required number of order $21$ elements all within some finite subgroup (and no other such elements in the group given by the presentation).
Please help :)
Here $varphi$ is Euler's totient function.
$dagger$ The Dunning-Kruger effect in action . . . It's the right number, yeah, but, as pointed out below, I didn't show that such a group actually exists.
group-theory finite-groups totient-function group-presentation
asked Jan 15 at 14:56
ShaunShaun
9,241113684
$endgroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.46.
Which of the following numbers could be the exact number of elements of order $21$ in a group: $21600, 21602, 21604$?
My Attempt:
Lemma: In a finite group, the number of elements of order $d$ is a multiple of $varphi(d)$.
(This is a Corollary of Theorem 4.4 ibid.)
Since $varphi(21)=12$ and $12mid 21600$ but not the other two candidate numbers, the lemma above implies that $21600$ is the exact number of elements of order $21$ for some (finite) group.
I'm quite sure I got this right$^{dagger}$. I intuited the result, having forgotten the Lemma (well, Corollary) above.
This question is just so I can make note of it, really, but I suppose I could use the post to ask the following:
Is the assumption that the group is finite necessary?
My guess is that it's not. I have a rough idea of using a presentation with some free generators in such a way that it has the required number of order $21$ elements all within some finite subgroup (and no other such elements in the group given by the presentation).
Please help :)
Here $varphi$ is Euler's totient function.
$dagger$ The Dunning-Kruger effect in action . . . It's the right number, yeah, but, as pointed out below, I didn't show that such a group actually exists.
group-theory finite-groups totient-function group-presentation
group-theory finite-groups totient-function group-presentation
asked Jan 15 at 14:56
ShaunShaun
9,241113684
asked Jan 15 at 14:56
ShaunShaun
9,241113684
edited Jan 23 at 21:08
Shaun
asked Jan 15 at 14:56
ShaunShaun
9,241113684
asked Jan 15 at 14:56
ShaunShaun
9,241113684
asked Jan 15 at 14:56
ShaunShaun
9,241113684
9,241113684
1
$begingroup$
I don't know if I'm being naive here, but if you had some infinite group $G$ and you were concerned about the possible precise number of elements of order $d$ that $G$ could have (given that the number of such elements is finite), could you not just consider $H$ to be the smallest subgroup of $G$ containing all elements of order $d$ (or even just the subgroup generated by the elements of order $d$) and then apply your lemma?
$endgroup$
– Adam Higgins
Jan 15 at 15:03
3
$begingroup$
@AdamHiggins $H$ is not necessarily finite: for example, if we take the group with presentation $langle a, b | a^{21}, b^{21}rangle$, then no finite (indeed, no proper) subgroup contains both $a$ and $b$.
$endgroup$
– user3482749
Jan 15 at 15:04
2
$begingroup$
@AdamHiggins $H$ might still be infinite.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 15:04
2
$begingroup$
Thanks all! My bad!
$endgroup$
– Adam Higgins
Jan 15 at 15:05
1
$begingroup$
You also need to show an example to a group such that the number of elements of order $21$ is exactly $21600$. The lemma you cited only shows that the other two possibilities are wrong, it does not give you that the remaining one is surely right.
$endgroup$
– A. Pongrácz
Jan 15 at 15:29
|
show 3 more comments
1
$begingroup$
I don't know if I'm being naive here, but if you had some infinite group $G$ and you were concerned about the possible precise number of elements of order $d$ that $G$ could have (given that the number of such elements is finite), could you not just consider $H$ to be the smallest subgroup of $G$ containing all elements of order $d$ (or even just the subgroup generated by the elements of order $d$) and then apply your lemma?
$endgroup$
– Adam Higgins
Jan 15 at 15:03
3
$begingroup$
@AdamHiggins $H$ is not necessarily finite: for example, if we take the group with presentation $langle a, b | a^{21}, b^{21}rangle$, then no finite (indeed, no proper) subgroup contains both $a$ and $b$.
$endgroup$
– user3482749
Jan 15 at 15:04
2
$begingroup$
@AdamHiggins $H$ might still be infinite.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 15:04
2
$begingroup$
Thanks all! My bad!
$endgroup$
– Adam Higgins
Jan 15 at 15:05
1
$begingroup$
You also need to show an example to a group such that the number of elements of order $21$ is exactly $21600$. The lemma you cited only shows that the other two possibilities are wrong, it does not give you that the remaining one is surely right.
$endgroup$
– A. Pongrácz
Jan 15 at 15:29
1
1
$begingroup$
I don't know if I'm being naive here, but if you had some infinite group $G$ and you were concerned about the possible precise number of elements of order $d$ that $G$ could have (given that the number of such elements is finite), could you not just consider $H$ to be the smallest subgroup of $G$ containing all elements of order $d$ (or even just the subgroup generated by the elements of order $d$) and then apply your lemma?
$endgroup$
– Adam Higgins
Jan 15 at 15:03
$begingroup$
I don't know if I'm being naive here, but if you had some infinite group $G$ and you were concerned about the possible precise number of elements of order $d$ that $G$ could have (given that the number of such elements is finite), could you not just consider $H$ to be the smallest subgroup of $G$ containing all elements of order $d$ (or even just the subgroup generated by the elements of order $d$) and then apply your lemma?
$endgroup$
– Adam Higgins
Jan 15 at 15:03
3
3
$begingroup$
@AdamHiggins $H$ is not necessarily finite: for example, if we take the group with presentation $langle a, b | a^{21}, b^{21}rangle$, then no finite (indeed, no proper) subgroup contains both $a$ and $b$.
$endgroup$
– user3482749
Jan 15 at 15:04
$begingroup$
@AdamHiggins $H$ is not necessarily finite: for example, if we take the group with presentation $langle a, b | a^{21}, b^{21}rangle$, then no finite (indeed, no proper) subgroup contains both $a$ and $b$.
$endgroup$
– user3482749
Jan 15 at 15:04
2
2
$begingroup$
@AdamHiggins $H$ might still be infinite.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 15:04
$begingroup$
@AdamHiggins $H$ might still be infinite.
$endgroup$
– Lord Shark the Unknown
Jan 15 at 15:04
2
2
$begingroup$
Thanks all! My bad!
$endgroup$
– Adam Higgins
Jan 15 at 15:05
$begingroup$
Thanks all! My bad!
$endgroup$
– Adam Higgins
Jan 15 at 15:05
1
1
$begingroup$
You also need to show an example to a group such that the number of elements of order $21$ is exactly $21600$. The lemma you cited only shows that the other two possibilities are wrong, it does not give you that the remaining one is surely right.
$endgroup$
– A. Pongrácz
Jan 15 at 15:29
$begingroup$
You also need to show an example to a group such that the number of elements of order $21$ is exactly $21600$. The lemma you cited only shows that the other two possibilities are wrong, it does not give you that the remaining one is surely right.
$endgroup$
– A. Pongrácz
Jan 15 at 15:29
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
There is no need to assume that $G$ is finite. For an arbitrary group $G$, and some fixed $n >0$, let $G_n$ be the set of elements of $G$ of (finite) order $n$. Then you can show that, for $g,h in G_n$, $g$ is a power of $h$ if and only if $h$ is a power of $g$, so "is a power of" is an equivalence relation on $G_n$, and all equivalence classes have size $phi(n)$.
So, in particular, if $G_n$ is finite, then $|G_n|$ is a multiple of $phi(n)$.
On the other hand, if $G_n$ is finite, then $C_G(G_n)$ has finite index in $G$, and I think that implies that $langle G_n rangle$ is finite.
answered Jan 15 at 15:04
Derek HoltDerek Holt
53.7k53571
$endgroup$
add a comment |
$begingroup$
The thing missing from your argument: there is a group such that there are exactly $21600$ elements of order $21$, and it is not so easy to find one.
Observe that $21600= 48cdot 450$.
In $Z_7times Z_7$, there are $48$ elements of order $7$.
In the nontrivial semidirect product $Z_{151}rtimes Z_3$, the possible orders of elements are $1,3$ and $151$, as $151$ is a prime, and the group is not cyclic. There is such a nontrivial semidirect product as $151equiv 1 pmod 3$.
Elements of order $1$ and $151$ are exactly those in the normal subgroup of order $151$. Thus there are exactly $3cdot 150= 450$ elements of order $3$ in $Z_{151}rtimes Z_3$.
Hence, in $(Z_7times Z_7)times (Z_{151}rtimes Z_3)$, there are exactly $21600$ elements of order $21$. (Think about it.)
answered Jan 15 at 15:48
A. PongráczA. Pongrácz
5,9531929
$endgroup$
1
$begingroup$
It depends on which one is the normal subgroup. In my case, the thing on the left is the normal subgroup, so I had to use $rtimes$. (The triangle points towards the normal subgroup.)
$endgroup$
– A. Pongrácz
Jan 15 at 15:53
$begingroup$
(Sorry: I deleted my last comment once I saw you'd edited your answer.) Yes, that's how I remember it too; I still get them mixed up every once in a while :)
$endgroup$
– Shaun
Jan 15 at 15:54
add a comment |
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2 Answers
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2 Answers
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$begingroup$
There is no need to assume that $G$ is finite. For an arbitrary group $G$, and some fixed $n >0$, let $G_n$ be the set of elements of $G$ of (finite) order $n$. Then you can show that, for $g,h in G_n$, $g$ is a power of $h$ if and only if $h$ is a power of $g$, so "is a power of" is an equivalence relation on $G_n$, and all equivalence classes have size $phi(n)$.
So, in particular, if $G_n$ is finite, then $|G_n|$ is a multiple of $phi(n)$.
On the other hand, if $G_n$ is finite, then $C_G(G_n)$ has finite index in $G$, and I think that implies that $langle G_n rangle$ is finite.
answered Jan 15 at 15:04
Derek HoltDerek Holt
53.7k53571
$endgroup$
add a comment |
$begingroup$
There is no need to assume that $G$ is finite. For an arbitrary group $G$, and some fixed $n >0$, let $G_n$ be the set of elements of $G$ of (finite) order $n$. Then you can show that, for $g,h in G_n$, $g$ is a power of $h$ if and only if $h$ is a power of $g$, so "is a power of" is an equivalence relation on $G_n$, and all equivalence classes have size $phi(n)$.
So, in particular, if $G_n$ is finite, then $|G_n|$ is a multiple of $phi(n)$.
On the other hand, if $G_n$ is finite, then $C_G(G_n)$ has finite index in $G$, and I think that implies that $langle G_n rangle$ is finite.
answered Jan 15 at 15:04
Derek HoltDerek Holt
53.7k53571
$endgroup$
add a comment |
$begingroup$
There is no need to assume that $G$ is finite. For an arbitrary group $G$, and some fixed $n >0$, let $G_n$ be the set of elements of $G$ of (finite) order $n$. Then you can show that, for $g,h in G_n$, $g$ is a power of $h$ if and only if $h$ is a power of $g$, so "is a power of" is an equivalence relation on $G_n$, and all equivalence classes have size $phi(n)$.
So, in particular, if $G_n$ is finite, then $|G_n|$ is a multiple of $phi(n)$.
On the other hand, if $G_n$ is finite, then $C_G(G_n)$ has finite index in $G$, and I think that implies that $langle G_n rangle$ is finite.
answered Jan 15 at 15:04
Derek HoltDerek Holt
53.7k53571
$endgroup$
There is no need to assume that $G$ is finite. For an arbitrary group $G$, and some fixed $n >0$, let $G_n$ be the set of elements of $G$ of (finite) order $n$. Then you can show that, for $g,h in G_n$, $g$ is a power of $h$ if and only if $h$ is a power of $g$, so "is a power of" is an equivalence relation on $G_n$, and all equivalence classes have size $phi(n)$.
So, in particular, if $G_n$ is finite, then $|G_n|$ is a multiple of $phi(n)$.
On the other hand, if $G_n$ is finite, then $C_G(G_n)$ has finite index in $G$, and I think that implies that $langle G_n rangle$ is finite.
answered Jan 15 at 15:04
Derek HoltDerek Holt
53.7k53571
edited Jan 15 at 15:12
answered Jan 15 at 15:04
Derek HoltDerek Holt
53.7k53571
answered Jan 15 at 15:04
Derek HoltDerek Holt
53.7k53571
answered Jan 15 at 15:04
Derek HoltDerek Holt
53.7k53571
53.7k53571
add a comment |
add a comment |
$begingroup$
The thing missing from your argument: there is a group such that there are exactly $21600$ elements of order $21$, and it is not so easy to find one.
Observe that $21600= 48cdot 450$.
In $Z_7times Z_7$, there are $48$ elements of order $7$.
In the nontrivial semidirect product $Z_{151}rtimes Z_3$, the possible orders of elements are $1,3$ and $151$, as $151$ is a prime, and the group is not cyclic. There is such a nontrivial semidirect product as $151equiv 1 pmod 3$.
Elements of order $1$ and $151$ are exactly those in the normal subgroup of order $151$. Thus there are exactly $3cdot 150= 450$ elements of order $3$ in $Z_{151}rtimes Z_3$.
Hence, in $(Z_7times Z_7)times (Z_{151}rtimes Z_3)$, there are exactly $21600$ elements of order $21$. (Think about it.)
answered Jan 15 at 15:48
A. PongráczA. Pongrácz
5,9531929
$endgroup$
1
$begingroup$
It depends on which one is the normal subgroup. In my case, the thing on the left is the normal subgroup, so I had to use $rtimes$. (The triangle points towards the normal subgroup.)
$endgroup$
– A. Pongrácz
Jan 15 at 15:53
$begingroup$
(Sorry: I deleted my last comment once I saw you'd edited your answer.) Yes, that's how I remember it too; I still get them mixed up every once in a while :)
$endgroup$
– Shaun
Jan 15 at 15:54
add a comment |
$begingroup$
The thing missing from your argument: there is a group such that there are exactly $21600$ elements of order $21$, and it is not so easy to find one.
Observe that $21600= 48cdot 450$.
In $Z_7times Z_7$, there are $48$ elements of order $7$.
In the nontrivial semidirect product $Z_{151}rtimes Z_3$, the possible orders of elements are $1,3$ and $151$, as $151$ is a prime, and the group is not cyclic. There is such a nontrivial semidirect product as $151equiv 1 pmod 3$.
Elements of order $1$ and $151$ are exactly those in the normal subgroup of order $151$. Thus there are exactly $3cdot 150= 450$ elements of order $3$ in $Z_{151}rtimes Z_3$.
Hence, in $(Z_7times Z_7)times (Z_{151}rtimes Z_3)$, there are exactly $21600$ elements of order $21$. (Think about it.)
answered Jan 15 at 15:48
A. PongráczA. Pongrácz
5,9531929
$endgroup$
1
$begingroup$
It depends on which one is the normal subgroup. In my case, the thing on the left is the normal subgroup, so I had to use $rtimes$. (The triangle points towards the normal subgroup.)
$endgroup$
– A. Pongrácz
Jan 15 at 15:53
$begingroup$
(Sorry: I deleted my last comment once I saw you'd edited your answer.) Yes, that's how I remember it too; I still get them mixed up every once in a while :)
$endgroup$
– Shaun
Jan 15 at 15:54
add a comment |
$begingroup$
The thing missing from your argument: there is a group such that there are exactly $21600$ elements of order $21$, and it is not so easy to find one.
Observe that $21600= 48cdot 450$.
In $Z_7times Z_7$, there are $48$ elements of order $7$.
In the nontrivial semidirect product $Z_{151}rtimes Z_3$, the possible orders of elements are $1,3$ and $151$, as $151$ is a prime, and the group is not cyclic. There is such a nontrivial semidirect product as $151equiv 1 pmod 3$.
Elements of order $1$ and $151$ are exactly those in the normal subgroup of order $151$. Thus there are exactly $3cdot 150= 450$ elements of order $3$ in $Z_{151}rtimes Z_3$.
Hence, in $(Z_7times Z_7)times (Z_{151}rtimes Z_3)$, there are exactly $21600$ elements of order $21$. (Think about it.)
answered Jan 15 at 15:48
A. PongráczA. Pongrácz
5,9531929
$endgroup$
The thing missing from your argument: there is a group such that there are exactly $21600$ elements of order $21$, and it is not so easy to find one.
Observe that $21600= 48cdot 450$.
In $Z_7times Z_7$, there are $48$ elements of order $7$.
In the nontrivial semidirect product $Z_{151}rtimes Z_3$, the possible orders of elements are $1,3$ and $151$, as $151$ is a prime, and the group is not cyclic. There is such a nontrivial semidirect product as $151equiv 1 pmod 3$.
Elements of order $1$ and $151$ are exactly those in the normal subgroup of order $151$. Thus there are exactly $3cdot 150= 450$ elements of order $3$ in $Z_{151}rtimes Z_3$.
Hence, in $(Z_7times Z_7)times (Z_{151}rtimes Z_3)$, there are exactly $21600$ elements of order $21$. (Think about it.)
answered Jan 15 at 15:48
A. PongráczA. Pongrácz
5,9531929
answered Jan 15 at 15:48
A. PongráczA. Pongrácz
5,9531929
answered Jan 15 at 15:48
A. PongráczA. Pongrácz
5,9531929
answered Jan 15 at 15:48
A. PongráczA. Pongrácz
5,9531929
5,9531929
1
$begingroup$
It depends on which one is the normal subgroup. In my case, the thing on the left is the normal subgroup, so I had to use $rtimes$. (The triangle points towards the normal subgroup.)
$endgroup$
– A. Pongrácz
Jan 15 at 15:53
$begingroup$
(Sorry: I deleted my last comment once I saw you'd edited your answer.) Yes, that's how I remember it too; I still get them mixed up every once in a while :)
$endgroup$
– Shaun
Jan 15 at 15:54
add a comment |
1
$begingroup$
It depends on which one is the normal subgroup. In my case, the thing on the left is the normal subgroup, so I had to use $rtimes$. (The triangle points towards the normal subgroup.)
$endgroup$
– A. Pongrácz
Jan 15 at 15:53
$begingroup$
(Sorry: I deleted my last comment once I saw you'd edited your answer.) Yes, that's how I remember it too; I still get them mixed up every once in a while :)
$endgroup$
– Shaun
Jan 15 at 15:54
1
1
$begingroup$
It depends on which one is the normal subgroup. In my case, the thing on the left is the normal subgroup, so I had to use $rtimes$. (The triangle points towards the normal subgroup.)
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– A. Pongrácz
Jan 15 at 15:53
$begingroup$
It depends on which one is the normal subgroup. In my case, the thing on the left is the normal subgroup, so I had to use $rtimes$. (The triangle points towards the normal subgroup.)
$endgroup$
– A. Pongrácz
Jan 15 at 15:53
$begingroup$
(Sorry: I deleted my last comment once I saw you'd edited your answer.) Yes, that's how I remember it too; I still get them mixed up every once in a while :)
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– Shaun
Jan 15 at 15:54
$begingroup$
(Sorry: I deleted my last comment once I saw you'd edited your answer.) Yes, that's how I remember it too; I still get them mixed up every once in a while :)
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– Shaun
Jan 15 at 15:54
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1
$begingroup$
I don't know if I'm being naive here, but if you had some infinite group $G$ and you were concerned about the possible precise number of elements of order $d$ that $G$ could have (given that the number of such elements is finite), could you not just consider $H$ to be the smallest subgroup of $G$ containing all elements of order $d$ (or even just the subgroup generated by the elements of order $d$) and then apply your lemma?
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– Adam Higgins
Jan 15 at 15:03
3
$begingroup$
@AdamHiggins $H$ is not necessarily finite: for example, if we take the group with presentation $langle a, b | a^{21}, b^{21}rangle$, then no finite (indeed, no proper) subgroup contains both $a$ and $b$.
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– user3482749
Jan 15 at 15:04
2
$begingroup$
@AdamHiggins $H$ might still be infinite.
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– Lord Shark the Unknown
Jan 15 at 15:04
2
$begingroup$
Thanks all! My bad!
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– Adam Higgins
Jan 15 at 15:05
1
$begingroup$
You also need to show an example to a group such that the number of elements of order $21$ is exactly $21600$. The lemma you cited only shows that the other two possibilities are wrong, it does not give you that the remaining one is surely right.
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– A. Pongrácz
Jan 15 at 15:29