How many elements of order 5 are there in a group of order 45?
$begingroup$
The Problem:
As the title asks: How many elements of order $5$ are there in a group of order $45$?
My Approach:
Let $G$ be a group of order $45$.
My first thought is to hit it with Sylow's Theorems, which tells me that $G$ has a normal subgroup of order $5$ (call it $N$) and a normal subgroup of order $9$. Clearly $4$ out of the $5$ elements in $N$ must be of order $5$ (since they must divide the order of $N$); and clearly none of these elements are in the normal subgroup of order $9$.
So, this means $G$ has at least $4$ elements of order $5$. Are there any more, though? The only thing I can think of is to investigate the possibility of a subgroup of order 15... but how would I know if $G$ has such a subgroup? And, if it did, how would I know whether it intersects $N$ trivially?
abstract-algebra group-theory
asked Jan 15 at 22:35
thisisourconcerndudethisisourconcerndude
1,1421225
$endgroup$
add a comment |
$begingroup$
The Problem:
As the title asks: How many elements of order $5$ are there in a group of order $45$?
My Approach:
Let $G$ be a group of order $45$.
My first thought is to hit it with Sylow's Theorems, which tells me that $G$ has a normal subgroup of order $5$ (call it $N$) and a normal subgroup of order $9$. Clearly $4$ out of the $5$ elements in $N$ must be of order $5$ (since they must divide the order of $N$); and clearly none of these elements are in the normal subgroup of order $9$.
So, this means $G$ has at least $4$ elements of order $5$. Are there any more, though? The only thing I can think of is to investigate the possibility of a subgroup of order 15... but how would I know if $G$ has such a subgroup? And, if it did, how would I know whether it intersects $N$ trivially?
abstract-algebra group-theory
asked Jan 15 at 22:35
thisisourconcerndudethisisourconcerndude
1,1421225
$endgroup$
$begingroup$
Why does Sylow tell you that there is a normal subgroup of order $5$? Remember, all the Sylow $p-$ groups are conjugate so this would mean that there was a unique Sylow $5-$ group. Note: you are correct in this case, but I think it needs a clearer argument.
$endgroup$
– lulu
Jan 15 at 22:41
$begingroup$
A subgroup of order $15$ is the product of the subgroup of order $5$ and a subgroup of order $3$, contained in the $3$-Sylow subgroup.
$endgroup$
– Bernard
Jan 15 at 22:43
$begingroup$
@lulu: the number of $5$-Sylow subgroups is congruent to $1bmod 5$ and it is a divisor of $9$.
$endgroup$
– Bernard
Jan 15 at 22:45
$begingroup$
@Bernard Are you saying that there is a subgroup of order 15 contained in a subgroup of order 9?
$endgroup$
– thisisourconcerndude
Jan 15 at 22:45
$begingroup$
I didn't check in detail, but I'd think that we might have a semi-direct product of the subgroup of order $5$ and a subgroup of order $3$ contained in the $3$-Sylow.
$endgroup$
– Bernard
Jan 15 at 22:52
add a comment |
$begingroup$
The Problem:
As the title asks: How many elements of order $5$ are there in a group of order $45$?
My Approach:
Let $G$ be a group of order $45$.
My first thought is to hit it with Sylow's Theorems, which tells me that $G$ has a normal subgroup of order $5$ (call it $N$) and a normal subgroup of order $9$. Clearly $4$ out of the $5$ elements in $N$ must be of order $5$ (since they must divide the order of $N$); and clearly none of these elements are in the normal subgroup of order $9$.
So, this means $G$ has at least $4$ elements of order $5$. Are there any more, though? The only thing I can think of is to investigate the possibility of a subgroup of order 15... but how would I know if $G$ has such a subgroup? And, if it did, how would I know whether it intersects $N$ trivially?
abstract-algebra group-theory
asked Jan 15 at 22:35
thisisourconcerndudethisisourconcerndude
1,1421225
$endgroup$
The Problem:
As the title asks: How many elements of order $5$ are there in a group of order $45$?
My Approach:
Let $G$ be a group of order $45$.
My first thought is to hit it with Sylow's Theorems, which tells me that $G$ has a normal subgroup of order $5$ (call it $N$) and a normal subgroup of order $9$. Clearly $4$ out of the $5$ elements in $N$ must be of order $5$ (since they must divide the order of $N$); and clearly none of these elements are in the normal subgroup of order $9$.
So, this means $G$ has at least $4$ elements of order $5$. Are there any more, though? The only thing I can think of is to investigate the possibility of a subgroup of order 15... but how would I know if $G$ has such a subgroup? And, if it did, how would I know whether it intersects $N$ trivially?
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 15 at 22:35
thisisourconcerndudethisisourconcerndude
1,1421225
asked Jan 15 at 22:35
thisisourconcerndudethisisourconcerndude
1,1421225
asked Jan 15 at 22:35
thisisourconcerndudethisisourconcerndude
1,1421225
asked Jan 15 at 22:35
thisisourconcerndudethisisourconcerndude
1,1421225
asked Jan 15 at 22:35
thisisourconcerndudethisisourconcerndude
1,1421225
1,1421225
$begingroup$
Why does Sylow tell you that there is a normal subgroup of order $5$? Remember, all the Sylow $p-$ groups are conjugate so this would mean that there was a unique Sylow $5-$ group. Note: you are correct in this case, but I think it needs a clearer argument.
$endgroup$
– lulu
Jan 15 at 22:41
$begingroup$
A subgroup of order $15$ is the product of the subgroup of order $5$ and a subgroup of order $3$, contained in the $3$-Sylow subgroup.
$endgroup$
– Bernard
Jan 15 at 22:43
$begingroup$
@lulu: the number of $5$-Sylow subgroups is congruent to $1bmod 5$ and it is a divisor of $9$.
$endgroup$
– Bernard
Jan 15 at 22:45
$begingroup$
@Bernard Are you saying that there is a subgroup of order 15 contained in a subgroup of order 9?
$endgroup$
– thisisourconcerndude
Jan 15 at 22:45
$begingroup$
I didn't check in detail, but I'd think that we might have a semi-direct product of the subgroup of order $5$ and a subgroup of order $3$ contained in the $3$-Sylow.
$endgroup$
– Bernard
Jan 15 at 22:52
add a comment |
$begingroup$
Why does Sylow tell you that there is a normal subgroup of order $5$? Remember, all the Sylow $p-$ groups are conjugate so this would mean that there was a unique Sylow $5-$ group. Note: you are correct in this case, but I think it needs a clearer argument.
$endgroup$
– lulu
Jan 15 at 22:41
$begingroup$
A subgroup of order $15$ is the product of the subgroup of order $5$ and a subgroup of order $3$, contained in the $3$-Sylow subgroup.
$endgroup$
– Bernard
Jan 15 at 22:43
$begingroup$
@lulu: the number of $5$-Sylow subgroups is congruent to $1bmod 5$ and it is a divisor of $9$.
$endgroup$
– Bernard
Jan 15 at 22:45
$begingroup$
@Bernard Are you saying that there is a subgroup of order 15 contained in a subgroup of order 9?
$endgroup$
– thisisourconcerndude
Jan 15 at 22:45
$begingroup$
I didn't check in detail, but I'd think that we might have a semi-direct product of the subgroup of order $5$ and a subgroup of order $3$ contained in the $3$-Sylow.
$endgroup$
– Bernard
Jan 15 at 22:52
$begingroup$
Why does Sylow tell you that there is a normal subgroup of order $5$? Remember, all the Sylow $p-$ groups are conjugate so this would mean that there was a unique Sylow $5-$ group. Note: you are correct in this case, but I think it needs a clearer argument.
$endgroup$
– lulu
Jan 15 at 22:41
$begingroup$
Why does Sylow tell you that there is a normal subgroup of order $5$? Remember, all the Sylow $p-$ groups are conjugate so this would mean that there was a unique Sylow $5-$ group. Note: you are correct in this case, but I think it needs a clearer argument.
$endgroup$
– lulu
Jan 15 at 22:41
$begingroup$
A subgroup of order $15$ is the product of the subgroup of order $5$ and a subgroup of order $3$, contained in the $3$-Sylow subgroup.
$endgroup$
– Bernard
Jan 15 at 22:43
$begingroup$
A subgroup of order $15$ is the product of the subgroup of order $5$ and a subgroup of order $3$, contained in the $3$-Sylow subgroup.
$endgroup$
– Bernard
Jan 15 at 22:43
$begingroup$
@lulu: the number of $5$-Sylow subgroups is congruent to $1bmod 5$ and it is a divisor of $9$.
$endgroup$
– Bernard
Jan 15 at 22:45
$begingroup$
@lulu: the number of $5$-Sylow subgroups is congruent to $1bmod 5$ and it is a divisor of $9$.
$endgroup$
– Bernard
Jan 15 at 22:45
$begingroup$
@Bernard Are you saying that there is a subgroup of order 15 contained in a subgroup of order 9?
$endgroup$
– thisisourconcerndude
Jan 15 at 22:45
$begingroup$
@Bernard Are you saying that there is a subgroup of order 15 contained in a subgroup of order 9?
$endgroup$
– thisisourconcerndude
Jan 15 at 22:45
$begingroup$
I didn't check in detail, but I'd think that we might have a semi-direct product of the subgroup of order $5$ and a subgroup of order $3$ contained in the $3$-Sylow.
$endgroup$
– Bernard
Jan 15 at 22:52
$begingroup$
I didn't check in detail, but I'd think that we might have a semi-direct product of the subgroup of order $5$ and a subgroup of order $3$ contained in the $3$-Sylow.
$endgroup$
– Bernard
Jan 15 at 22:52
add a comment |
1 Answer
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$begingroup$
You already did 70% of the work:
Let $G$ be of order $45 = 5cdot 3^2$. By Sylow's second theorem, the number of 5-sylow-groups $n_5$ must be either 1, 3, or 9. Out of those choices, only $n_5=1equiv 1$ modulo 5. Therefore, only one subgroup of order 5 exists (normality isn't even important here).
Now note that every possible element of order five generates a subgroup of order five (in which it must be contained). I'm confident you can take it from here. :-)
answered Jan 15 at 22:51
LukeLuke
1,2391026
$endgroup$
$begingroup$
So, if $x in G$ with $|x| = 5$, then $langle x rangle$ is a subgroup of $G$ with $|langle x rangle| = 5$. Since $|langle x rangle| = 5$, it must be conjugate to $N$; but since there's only one Sylow 5-subgroup (i.e., $N$), it must be that $langle x rangle = N$. Therefore, there must be exactly 4 elements of order 5 in $G$. Right?
$endgroup$
– thisisourconcerndude
Jan 16 at 15:39
$begingroup$
That is correct, although the fact that every Sylow-p-groups are conjugate isn't even necessary: every group of order five is, by definition, a Sylow-5-subgroup of $G$. Therefore, given we only have one Sylow-5-subgroup, this is precisely what $langle x rangle$ must be.
$endgroup$
– Luke
Jan 16 at 19:07
add a comment |
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1 Answer
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$begingroup$
You already did 70% of the work:
Let $G$ be of order $45 = 5cdot 3^2$. By Sylow's second theorem, the number of 5-sylow-groups $n_5$ must be either 1, 3, or 9. Out of those choices, only $n_5=1equiv 1$ modulo 5. Therefore, only one subgroup of order 5 exists (normality isn't even important here).
Now note that every possible element of order five generates a subgroup of order five (in which it must be contained). I'm confident you can take it from here. :-)
answered Jan 15 at 22:51
LukeLuke
1,2391026
$endgroup$
$begingroup$
So, if $x in G$ with $|x| = 5$, then $langle x rangle$ is a subgroup of $G$ with $|langle x rangle| = 5$. Since $|langle x rangle| = 5$, it must be conjugate to $N$; but since there's only one Sylow 5-subgroup (i.e., $N$), it must be that $langle x rangle = N$. Therefore, there must be exactly 4 elements of order 5 in $G$. Right?
$endgroup$
– thisisourconcerndude
Jan 16 at 15:39
$begingroup$
That is correct, although the fact that every Sylow-p-groups are conjugate isn't even necessary: every group of order five is, by definition, a Sylow-5-subgroup of $G$. Therefore, given we only have one Sylow-5-subgroup, this is precisely what $langle x rangle$ must be.
$endgroup$
– Luke
Jan 16 at 19:07
add a comment |
$begingroup$
You already did 70% of the work:
Let $G$ be of order $45 = 5cdot 3^2$. By Sylow's second theorem, the number of 5-sylow-groups $n_5$ must be either 1, 3, or 9. Out of those choices, only $n_5=1equiv 1$ modulo 5. Therefore, only one subgroup of order 5 exists (normality isn't even important here).
Now note that every possible element of order five generates a subgroup of order five (in which it must be contained). I'm confident you can take it from here. :-)
answered Jan 15 at 22:51
LukeLuke
1,2391026
$endgroup$
$begingroup$
So, if $x in G$ with $|x| = 5$, then $langle x rangle$ is a subgroup of $G$ with $|langle x rangle| = 5$. Since $|langle x rangle| = 5$, it must be conjugate to $N$; but since there's only one Sylow 5-subgroup (i.e., $N$), it must be that $langle x rangle = N$. Therefore, there must be exactly 4 elements of order 5 in $G$. Right?
$endgroup$
– thisisourconcerndude
Jan 16 at 15:39
$begingroup$
That is correct, although the fact that every Sylow-p-groups are conjugate isn't even necessary: every group of order five is, by definition, a Sylow-5-subgroup of $G$. Therefore, given we only have one Sylow-5-subgroup, this is precisely what $langle x rangle$ must be.
$endgroup$
– Luke
Jan 16 at 19:07
add a comment |
$begingroup$
You already did 70% of the work:
Let $G$ be of order $45 = 5cdot 3^2$. By Sylow's second theorem, the number of 5-sylow-groups $n_5$ must be either 1, 3, or 9. Out of those choices, only $n_5=1equiv 1$ modulo 5. Therefore, only one subgroup of order 5 exists (normality isn't even important here).
Now note that every possible element of order five generates a subgroup of order five (in which it must be contained). I'm confident you can take it from here. :-)
answered Jan 15 at 22:51
LukeLuke
1,2391026
$endgroup$
You already did 70% of the work:
Let $G$ be of order $45 = 5cdot 3^2$. By Sylow's second theorem, the number of 5-sylow-groups $n_5$ must be either 1, 3, or 9. Out of those choices, only $n_5=1equiv 1$ modulo 5. Therefore, only one subgroup of order 5 exists (normality isn't even important here).
Now note that every possible element of order five generates a subgroup of order five (in which it must be contained). I'm confident you can take it from here. :-)
answered Jan 15 at 22:51
LukeLuke
1,2391026
answered Jan 15 at 22:51
LukeLuke
1,2391026
answered Jan 15 at 22:51
LukeLuke
1,2391026
answered Jan 15 at 22:51
LukeLuke
1,2391026
1,2391026
$begingroup$
So, if $x in G$ with $|x| = 5$, then $langle x rangle$ is a subgroup of $G$ with $|langle x rangle| = 5$. Since $|langle x rangle| = 5$, it must be conjugate to $N$; but since there's only one Sylow 5-subgroup (i.e., $N$), it must be that $langle x rangle = N$. Therefore, there must be exactly 4 elements of order 5 in $G$. Right?
$endgroup$
– thisisourconcerndude
Jan 16 at 15:39
$begingroup$
That is correct, although the fact that every Sylow-p-groups are conjugate isn't even necessary: every group of order five is, by definition, a Sylow-5-subgroup of $G$. Therefore, given we only have one Sylow-5-subgroup, this is precisely what $langle x rangle$ must be.
$endgroup$
– Luke
Jan 16 at 19:07
add a comment |
$begingroup$
So, if $x in G$ with $|x| = 5$, then $langle x rangle$ is a subgroup of $G$ with $|langle x rangle| = 5$. Since $|langle x rangle| = 5$, it must be conjugate to $N$; but since there's only one Sylow 5-subgroup (i.e., $N$), it must be that $langle x rangle = N$. Therefore, there must be exactly 4 elements of order 5 in $G$. Right?
$endgroup$
– thisisourconcerndude
Jan 16 at 15:39
$begingroup$
That is correct, although the fact that every Sylow-p-groups are conjugate isn't even necessary: every group of order five is, by definition, a Sylow-5-subgroup of $G$. Therefore, given we only have one Sylow-5-subgroup, this is precisely what $langle x rangle$ must be.
$endgroup$
– Luke
Jan 16 at 19:07
$begingroup$
So, if $x in G$ with $|x| = 5$, then $langle x rangle$ is a subgroup of $G$ with $|langle x rangle| = 5$. Since $|langle x rangle| = 5$, it must be conjugate to $N$; but since there's only one Sylow 5-subgroup (i.e., $N$), it must be that $langle x rangle = N$. Therefore, there must be exactly 4 elements of order 5 in $G$. Right?
$endgroup$
– thisisourconcerndude
Jan 16 at 15:39
$begingroup$
So, if $x in G$ with $|x| = 5$, then $langle x rangle$ is a subgroup of $G$ with $|langle x rangle| = 5$. Since $|langle x rangle| = 5$, it must be conjugate to $N$; but since there's only one Sylow 5-subgroup (i.e., $N$), it must be that $langle x rangle = N$. Therefore, there must be exactly 4 elements of order 5 in $G$. Right?
$endgroup$
– thisisourconcerndude
Jan 16 at 15:39
$begingroup$
That is correct, although the fact that every Sylow-p-groups are conjugate isn't even necessary: every group of order five is, by definition, a Sylow-5-subgroup of $G$. Therefore, given we only have one Sylow-5-subgroup, this is precisely what $langle x rangle$ must be.
$endgroup$
– Luke
Jan 16 at 19:07
$begingroup$
That is correct, although the fact that every Sylow-p-groups are conjugate isn't even necessary: every group of order five is, by definition, a Sylow-5-subgroup of $G$. Therefore, given we only have one Sylow-5-subgroup, this is precisely what $langle x rangle$ must be.
$endgroup$
– Luke
Jan 16 at 19:07
add a comment |
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$begingroup$
Why does Sylow tell you that there is a normal subgroup of order $5$? Remember, all the Sylow $p-$ groups are conjugate so this would mean that there was a unique Sylow $5-$ group. Note: you are correct in this case, but I think it needs a clearer argument.
$endgroup$
– lulu
Jan 15 at 22:41
$begingroup$
A subgroup of order $15$ is the product of the subgroup of order $5$ and a subgroup of order $3$, contained in the $3$-Sylow subgroup.
$endgroup$
– Bernard
Jan 15 at 22:43
$begingroup$
@lulu: the number of $5$-Sylow subgroups is congruent to $1bmod 5$ and it is a divisor of $9$.
$endgroup$
– Bernard
Jan 15 at 22:45
$begingroup$
@Bernard Are you saying that there is a subgroup of order 15 contained in a subgroup of order 9?
$endgroup$
– thisisourconcerndude
Jan 15 at 22:45
$begingroup$
I didn't check in detail, but I'd think that we might have a semi-direct product of the subgroup of order $5$ and a subgroup of order $3$ contained in the $3$-Sylow.
$endgroup$
– Bernard
Jan 15 at 22:52