descriptions of groups of order $45$
$begingroup$
Describe the structure of groups of order $45$
Let $|G|=45$
If $G$ has an element of order $45$
$G=＜x|x^{45}=1＞$
If $G$ doesn’t have an element of order $45$
$P_9$: $9$-Sylow group
$P_5$: $5$-Sylow group
$P_9 triangleleft G$, $P_5 triangleleft G$, $P_9 cap P_5={1}$
Hence $P_9×P_5$ (direct)
$G=P_9×P_5$ (I’m not sure about this)
$P_9=＜x,y| x^3=y^3=1, xy=yx＞$
$P_5=＜x| x^5=1＞$
This is what I’ve got.
I can’t describe the structure of $P_9×P_5$
I want to describe it like $G=＜...＞$
group-theory
edited Jan 20 at 1:10
user289143
903313
asked Jan 20 at 0:22
sakisaki
337
$endgroup$
add a comment |
$begingroup$
Describe the structure of groups of order $45$
Let $|G|=45$
If $G$ has an element of order $45$
$G=＜x|x^{45}=1＞$
If $G$ doesn’t have an element of order $45$
$P_9$: $9$-Sylow group
$P_5$: $5$-Sylow group
$P_9 triangleleft G$, $P_5 triangleleft G$, $P_9 cap P_5={1}$
Hence $P_9×P_5$ (direct)
$G=P_9×P_5$ (I’m not sure about this)
$P_9=＜x,y| x^3=y^3=1, xy=yx＞$
$P_5=＜x| x^5=1＞$
This is what I’ve got.
I can’t describe the structure of $P_9×P_5$
I want to describe it like $G=＜...＞$
group-theory
edited Jan 20 at 1:10
user289143
903313
asked Jan 20 at 0:22
sakisaki
337
$endgroup$
add a comment |
$begingroup$
Describe the structure of groups of order $45$
Let $|G|=45$
If $G$ has an element of order $45$
$G=＜x|x^{45}=1＞$
If $G$ doesn’t have an element of order $45$
$P_9$: $9$-Sylow group
$P_5$: $5$-Sylow group
$P_9 triangleleft G$, $P_5 triangleleft G$, $P_9 cap P_5={1}$
Hence $P_9×P_5$ (direct)
$G=P_9×P_5$ (I’m not sure about this)
$P_9=＜x,y| x^3=y^3=1, xy=yx＞$
$P_5=＜x| x^5=1＞$
This is what I’ve got.
I can’t describe the structure of $P_9×P_5$
I want to describe it like $G=＜...＞$
group-theory
edited Jan 20 at 1:10
user289143
903313
asked Jan 20 at 0:22
sakisaki
337
$endgroup$
Describe the structure of groups of order $45$
Let $|G|=45$
If $G$ has an element of order $45$
$G=＜x|x^{45}=1＞$
If $G$ doesn’t have an element of order $45$
$P_9$: $9$-Sylow group
$P_5$: $5$-Sylow group
$P_9 triangleleft G$, $P_5 triangleleft G$, $P_9 cap P_5={1}$
Hence $P_9×P_5$ (direct)
$G=P_9×P_5$ (I’m not sure about this)
$P_9=＜x,y| x^3=y^3=1, xy=yx＞$
$P_5=＜x| x^5=1＞$
This is what I’ve got.
I can’t describe the structure of $P_9×P_5$
I want to describe it like $G=＜...＞$
group-theory
group-theory
edited Jan 20 at 1:10
user289143
903313
asked Jan 20 at 0:22
sakisaki
337
edited Jan 20 at 1:10
user289143
903313
asked Jan 20 at 0:22
sakisaki
337
edited Jan 20 at 1:10
user289143
903313
edited Jan 20 at 1:10
user289143
903313
edited Jan 20 at 1:10
user289143
903313
903313
asked Jan 20 at 0:22
sakisaki
337
asked Jan 20 at 0:22
sakisaki
337
asked Jan 20 at 0:22
sakisaki
337
337
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your argument is fine. In a group of order $45$, the Sylow $3$-subgroup and the Sylow $5$-subgroup are necessarily normal. Hence the group is the direct product of these subgroups.
For a group of order $9$, you have two possibilities :
Either $C_9 = langle xmid x^9=1rangle$ or $C_3times C_3=langle x,ymid x^3=y^3=[x,y]=1rangle$.
There is only one group of order $5$, namely $C_5=langle xmid x^5=1rangle$.
Hence, there are two groups of order $45$, namely $C_9times C_5=C_{45}=langle xmid x^{45}=1rangle$ or $C_3times C_3times C_5=langle x,y,zmid x^3=y^3=z^5=[x,y]=[x,z]=[y,z]=1rangle$.
answered Jan 20 at 1:02
LeventLevent
2,729925
$endgroup$
$begingroup$
Thank you very much! And both of these groups are abelian, right?
$endgroup$
– saki
Jan 20 at 1:20
$begingroup$
Yes, they are both abelian.
$endgroup$
– Levent
Jan 20 at 1:21
$begingroup$
I inderstand. Thank you!
$endgroup$
– saki
Jan 21 at 1:39
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
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active
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active
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$begingroup$
Your argument is fine. In a group of order $45$, the Sylow $3$-subgroup and the Sylow $5$-subgroup are necessarily normal. Hence the group is the direct product of these subgroups.
For a group of order $9$, you have two possibilities :
Either $C_9 = langle xmid x^9=1rangle$ or $C_3times C_3=langle x,ymid x^3=y^3=[x,y]=1rangle$.
There is only one group of order $5$, namely $C_5=langle xmid x^5=1rangle$.
Hence, there are two groups of order $45$, namely $C_9times C_5=C_{45}=langle xmid x^{45}=1rangle$ or $C_3times C_3times C_5=langle x,y,zmid x^3=y^3=z^5=[x,y]=[x,z]=[y,z]=1rangle$.
answered Jan 20 at 1:02
LeventLevent
2,729925
$endgroup$
$begingroup$
Thank you very much! And both of these groups are abelian, right?
$endgroup$
– saki
Jan 20 at 1:20
$begingroup$
Yes, they are both abelian.
$endgroup$
– Levent
Jan 20 at 1:21
$begingroup$
I inderstand. Thank you!
$endgroup$
– saki
Jan 21 at 1:39
add a comment |
$begingroup$
Your argument is fine. In a group of order $45$, the Sylow $3$-subgroup and the Sylow $5$-subgroup are necessarily normal. Hence the group is the direct product of these subgroups.
For a group of order $9$, you have two possibilities :
Either $C_9 = langle xmid x^9=1rangle$ or $C_3times C_3=langle x,ymid x^3=y^3=[x,y]=1rangle$.
There is only one group of order $5$, namely $C_5=langle xmid x^5=1rangle$.
Hence, there are two groups of order $45$, namely $C_9times C_5=C_{45}=langle xmid x^{45}=1rangle$ or $C_3times C_3times C_5=langle x,y,zmid x^3=y^3=z^5=[x,y]=[x,z]=[y,z]=1rangle$.
answered Jan 20 at 1:02
LeventLevent
2,729925
$endgroup$
$begingroup$
Thank you very much! And both of these groups are abelian, right?
$endgroup$
– saki
Jan 20 at 1:20
$begingroup$
Yes, they are both abelian.
$endgroup$
– Levent
Jan 20 at 1:21
$begingroup$
I inderstand. Thank you!
$endgroup$
– saki
Jan 21 at 1:39
add a comment |
$begingroup$
Your argument is fine. In a group of order $45$, the Sylow $3$-subgroup and the Sylow $5$-subgroup are necessarily normal. Hence the group is the direct product of these subgroups.
For a group of order $9$, you have two possibilities :
Either $C_9 = langle xmid x^9=1rangle$ or $C_3times C_3=langle x,ymid x^3=y^3=[x,y]=1rangle$.
There is only one group of order $5$, namely $C_5=langle xmid x^5=1rangle$.
Hence, there are two groups of order $45$, namely $C_9times C_5=C_{45}=langle xmid x^{45}=1rangle$ or $C_3times C_3times C_5=langle x,y,zmid x^3=y^3=z^5=[x,y]=[x,z]=[y,z]=1rangle$.
answered Jan 20 at 1:02
LeventLevent
2,729925
$endgroup$
Your argument is fine. In a group of order $45$, the Sylow $3$-subgroup and the Sylow $5$-subgroup are necessarily normal. Hence the group is the direct product of these subgroups.
For a group of order $9$, you have two possibilities :
Either $C_9 = langle xmid x^9=1rangle$ or $C_3times C_3=langle x,ymid x^3=y^3=[x,y]=1rangle$.
There is only one group of order $5$, namely $C_5=langle xmid x^5=1rangle$.
Hence, there are two groups of order $45$, namely $C_9times C_5=C_{45}=langle xmid x^{45}=1rangle$ or $C_3times C_3times C_5=langle x,y,zmid x^3=y^3=z^5=[x,y]=[x,z]=[y,z]=1rangle$.
answered Jan 20 at 1:02
LeventLevent
2,729925
answered Jan 20 at 1:02
LeventLevent
2,729925
answered Jan 20 at 1:02
LeventLevent
2,729925
answered Jan 20 at 1:02
LeventLevent
2,729925
2,729925
$begingroup$
Thank you very much! And both of these groups are abelian, right?
$endgroup$
– saki
Jan 20 at 1:20
$begingroup$
Yes, they are both abelian.
$endgroup$
– Levent
Jan 20 at 1:21
$begingroup$
I inderstand. Thank you!
$endgroup$
– saki
Jan 21 at 1:39
add a comment |
$begingroup$
Thank you very much! And both of these groups are abelian, right?
$endgroup$
– saki
Jan 20 at 1:20
$begingroup$
Yes, they are both abelian.
$endgroup$
– Levent
Jan 20 at 1:21
$begingroup$
I inderstand. Thank you!
$endgroup$
– saki
Jan 21 at 1:39
$begingroup$
Thank you very much! And both of these groups are abelian, right?
$endgroup$
– saki
Jan 20 at 1:20
$begingroup$
Thank you very much! And both of these groups are abelian, right?
$endgroup$
– saki
Jan 20 at 1:20
$begingroup$
Yes, they are both abelian.
$endgroup$
– Levent
Jan 20 at 1:21
$begingroup$
Yes, they are both abelian.
$endgroup$
– Levent
Jan 20 at 1:21
$begingroup$
I inderstand. Thank you!
$endgroup$
– saki
Jan 21 at 1:39
$begingroup$
I inderstand. Thank you!
$endgroup$
– saki
Jan 21 at 1:39
add a comment |
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