Is $A_4 times mathbb{Z}_3$ isomorphic to $S_3 times S_3$?
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Ιs $A_4 times mathbb{Z}_3$ isomorphic to $S_3 times S_3$?
I am trying to find an element of $S_3 times S_3$ which has an order, let's say $a$, and $A_4 times mathbb{Z}_3$ has no element of such order.
abstract-algebra group-theory finite-groups symmetric-groups
$endgroup$
add a comment |
$begingroup$
Ιs $A_4 times mathbb{Z}_3$ isomorphic to $S_3 times S_3$?
I am trying to find an element of $S_3 times S_3$ which has an order, let's say $a$, and $A_4 times mathbb{Z}_3$ has no element of such order.
abstract-algebra group-theory finite-groups symmetric-groups
$endgroup$
$begingroup$
You could count the elements of order $3$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:47
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Both groups have elements of orders $1$, $2$, $3$ and $6$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:47
$begingroup$
Can $S_3times S_3$ have a normal subgroup isomorphic to $A_4$?
$endgroup$
– Hagen von Eitzen
Jan 18 at 7:50
add a comment |
$begingroup$
Ιs $A_4 times mathbb{Z}_3$ isomorphic to $S_3 times S_3$?
I am trying to find an element of $S_3 times S_3$ which has an order, let's say $a$, and $A_4 times mathbb{Z}_3$ has no element of such order.
abstract-algebra group-theory finite-groups symmetric-groups
$endgroup$
Ιs $A_4 times mathbb{Z}_3$ isomorphic to $S_3 times S_3$?
I am trying to find an element of $S_3 times S_3$ which has an order, let's say $a$, and $A_4 times mathbb{Z}_3$ has no element of such order.
abstract-algebra group-theory finite-groups symmetric-groups
abstract-algebra group-theory finite-groups symmetric-groups
edited Jan 22 at 20:45
user26857
39.6k124284
39.6k124284
asked Jan 18 at 7:45
kotsos sgouraskotsos sgouras
157
157
$begingroup$
You could count the elements of order $3$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:47
$begingroup$
Both groups have elements of orders $1$, $2$, $3$ and $6$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:47
$begingroup$
Can $S_3times S_3$ have a normal subgroup isomorphic to $A_4$?
$endgroup$
– Hagen von Eitzen
Jan 18 at 7:50
add a comment |
$begingroup$
You could count the elements of order $3$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:47
$begingroup$
Both groups have elements of orders $1$, $2$, $3$ and $6$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:47
$begingroup$
Can $S_3times S_3$ have a normal subgroup isomorphic to $A_4$?
$endgroup$
– Hagen von Eitzen
Jan 18 at 7:50
$begingroup$
You could count the elements of order $3$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:47
$begingroup$
You could count the elements of order $3$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:47
$begingroup$
Both groups have elements of orders $1$, $2$, $3$ and $6$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:47
$begingroup$
Both groups have elements of orders $1$, $2$, $3$ and $6$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:47
$begingroup$
Can $S_3times S_3$ have a normal subgroup isomorphic to $A_4$?
$endgroup$
– Hagen von Eitzen
Jan 18 at 7:50
$begingroup$
Can $S_3times S_3$ have a normal subgroup isomorphic to $A_4$?
$endgroup$
– Hagen von Eitzen
Jan 18 at 7:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In $S_3times S_3$ there are
$1$ element of order $1$
$color{red}{15}$ elements of order $2$, $(tau_1,e),(e,tau_2),(tau_1,tau_2)$
$8$ elements of order $3$
$12$ elements of order $6$
while in $A_4timesmathbb{Z}_3$ there are
$1$ element of order $1$
$color{red}{3}$ elements of order $2$, $(tau_1 tau_2,e)$
$26$ elements of order $3$
$6$ elements of order $6$
so the easiest way to state $S_3times S_3 notsimeq A_4timesmathbb{Z}_3$ is probably to compare the number of involutions (elements with order $2$) in both groups. Or wonder about the largest abelian subgroups, which in the former case are isomorphic to $mathbb{Z}_3timesmathbb{Z}_3$ and in the latter case are isomorphic to $mathbb{Z}_2timesmathbb{Z}_2timesmathbb{Z}_3$.
$endgroup$
$begingroup$
$A_4$ contains only 3 elements of order two. I actually found 3 elements of order two, 26 of order three, and 6 of order six in $A_4timesmathbb{Z}_3$.
$endgroup$
– user26857
Jan 22 at 9:29
1
$begingroup$
@user26857: you're right, answer fixed.
$endgroup$
– Jack D'Aurizio
Jan 22 at 16:23
add a comment |
$begingroup$
You could use the following useful facts.
- Given $A$ and $B$ groups, $Z(A times B) = Z(A) times Z(B)$.
$Z(S_n)=lbrace id rbrace$ for all $n geq 3$.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
In $S_3times S_3$ there are
$1$ element of order $1$
$color{red}{15}$ elements of order $2$, $(tau_1,e),(e,tau_2),(tau_1,tau_2)$
$8$ elements of order $3$
$12$ elements of order $6$
while in $A_4timesmathbb{Z}_3$ there are
$1$ element of order $1$
$color{red}{3}$ elements of order $2$, $(tau_1 tau_2,e)$
$26$ elements of order $3$
$6$ elements of order $6$
so the easiest way to state $S_3times S_3 notsimeq A_4timesmathbb{Z}_3$ is probably to compare the number of involutions (elements with order $2$) in both groups. Or wonder about the largest abelian subgroups, which in the former case are isomorphic to $mathbb{Z}_3timesmathbb{Z}_3$ and in the latter case are isomorphic to $mathbb{Z}_2timesmathbb{Z}_2timesmathbb{Z}_3$.
$endgroup$
$begingroup$
$A_4$ contains only 3 elements of order two. I actually found 3 elements of order two, 26 of order three, and 6 of order six in $A_4timesmathbb{Z}_3$.
$endgroup$
– user26857
Jan 22 at 9:29
1
$begingroup$
@user26857: you're right, answer fixed.
$endgroup$
– Jack D'Aurizio
Jan 22 at 16:23
add a comment |
$begingroup$
In $S_3times S_3$ there are
$1$ element of order $1$
$color{red}{15}$ elements of order $2$, $(tau_1,e),(e,tau_2),(tau_1,tau_2)$
$8$ elements of order $3$
$12$ elements of order $6$
while in $A_4timesmathbb{Z}_3$ there are
$1$ element of order $1$
$color{red}{3}$ elements of order $2$, $(tau_1 tau_2,e)$
$26$ elements of order $3$
$6$ elements of order $6$
so the easiest way to state $S_3times S_3 notsimeq A_4timesmathbb{Z}_3$ is probably to compare the number of involutions (elements with order $2$) in both groups. Or wonder about the largest abelian subgroups, which in the former case are isomorphic to $mathbb{Z}_3timesmathbb{Z}_3$ and in the latter case are isomorphic to $mathbb{Z}_2timesmathbb{Z}_2timesmathbb{Z}_3$.
$endgroup$
$begingroup$
$A_4$ contains only 3 elements of order two. I actually found 3 elements of order two, 26 of order three, and 6 of order six in $A_4timesmathbb{Z}_3$.
$endgroup$
– user26857
Jan 22 at 9:29
1
$begingroup$
@user26857: you're right, answer fixed.
$endgroup$
– Jack D'Aurizio
Jan 22 at 16:23
add a comment |
$begingroup$
In $S_3times S_3$ there are
$1$ element of order $1$
$color{red}{15}$ elements of order $2$, $(tau_1,e),(e,tau_2),(tau_1,tau_2)$
$8$ elements of order $3$
$12$ elements of order $6$
while in $A_4timesmathbb{Z}_3$ there are
$1$ element of order $1$
$color{red}{3}$ elements of order $2$, $(tau_1 tau_2,e)$
$26$ elements of order $3$
$6$ elements of order $6$
so the easiest way to state $S_3times S_3 notsimeq A_4timesmathbb{Z}_3$ is probably to compare the number of involutions (elements with order $2$) in both groups. Or wonder about the largest abelian subgroups, which in the former case are isomorphic to $mathbb{Z}_3timesmathbb{Z}_3$ and in the latter case are isomorphic to $mathbb{Z}_2timesmathbb{Z}_2timesmathbb{Z}_3$.
$endgroup$
In $S_3times S_3$ there are
$1$ element of order $1$
$color{red}{15}$ elements of order $2$, $(tau_1,e),(e,tau_2),(tau_1,tau_2)$
$8$ elements of order $3$
$12$ elements of order $6$
while in $A_4timesmathbb{Z}_3$ there are
$1$ element of order $1$
$color{red}{3}$ elements of order $2$, $(tau_1 tau_2,e)$
$26$ elements of order $3$
$6$ elements of order $6$
so the easiest way to state $S_3times S_3 notsimeq A_4timesmathbb{Z}_3$ is probably to compare the number of involutions (elements with order $2$) in both groups. Or wonder about the largest abelian subgroups, which in the former case are isomorphic to $mathbb{Z}_3timesmathbb{Z}_3$ and in the latter case are isomorphic to $mathbb{Z}_2timesmathbb{Z}_2timesmathbb{Z}_3$.
edited Jan 22 at 16:23
answered Jan 18 at 9:02
Jack D'AurizioJack D'Aurizio
292k33284674
292k33284674
$begingroup$
$A_4$ contains only 3 elements of order two. I actually found 3 elements of order two, 26 of order three, and 6 of order six in $A_4timesmathbb{Z}_3$.
$endgroup$
– user26857
Jan 22 at 9:29
1
$begingroup$
@user26857: you're right, answer fixed.
$endgroup$
– Jack D'Aurizio
Jan 22 at 16:23
add a comment |
$begingroup$
$A_4$ contains only 3 elements of order two. I actually found 3 elements of order two, 26 of order three, and 6 of order six in $A_4timesmathbb{Z}_3$.
$endgroup$
– user26857
Jan 22 at 9:29
1
$begingroup$
@user26857: you're right, answer fixed.
$endgroup$
– Jack D'Aurizio
Jan 22 at 16:23
$begingroup$
$A_4$ contains only 3 elements of order two. I actually found 3 elements of order two, 26 of order three, and 6 of order six in $A_4timesmathbb{Z}_3$.
$endgroup$
– user26857
Jan 22 at 9:29
$begingroup$
$A_4$ contains only 3 elements of order two. I actually found 3 elements of order two, 26 of order three, and 6 of order six in $A_4timesmathbb{Z}_3$.
$endgroup$
– user26857
Jan 22 at 9:29
1
1
$begingroup$
@user26857: you're right, answer fixed.
$endgroup$
– Jack D'Aurizio
Jan 22 at 16:23
$begingroup$
@user26857: you're right, answer fixed.
$endgroup$
– Jack D'Aurizio
Jan 22 at 16:23
add a comment |
$begingroup$
You could use the following useful facts.
- Given $A$ and $B$ groups, $Z(A times B) = Z(A) times Z(B)$.
$Z(S_n)=lbrace id rbrace$ for all $n geq 3$.
$endgroup$
add a comment |
$begingroup$
You could use the following useful facts.
- Given $A$ and $B$ groups, $Z(A times B) = Z(A) times Z(B)$.
$Z(S_n)=lbrace id rbrace$ for all $n geq 3$.
$endgroup$
add a comment |
$begingroup$
You could use the following useful facts.
- Given $A$ and $B$ groups, $Z(A times B) = Z(A) times Z(B)$.
$Z(S_n)=lbrace id rbrace$ for all $n geq 3$.
$endgroup$
You could use the following useful facts.
- Given $A$ and $B$ groups, $Z(A times B) = Z(A) times Z(B)$.
$Z(S_n)=lbrace id rbrace$ for all $n geq 3$.
answered Jan 18 at 14:30
Pietro GheriPietro Gheri
1662
1662
add a comment |
add a comment |
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$begingroup$
You could count the elements of order $3$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:47
$begingroup$
Both groups have elements of orders $1$, $2$, $3$ and $6$.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 7:47
$begingroup$
Can $S_3times S_3$ have a normal subgroup isomorphic to $A_4$?
$endgroup$
– Hagen von Eitzen
Jan 18 at 7:50