Mixing
$D_3oplus D_4$ not isomorphic to $D_{24}$
We need to prove that $D_{3} oplus D_{4}$ is not isomorphic to $D_{24}$ .
The way in which I approach such type of questions is to count the number of elements of order $x$ in one group and then in the other group, and then conclude that they aren't equal and hence there can't be any isomorphism between them.
But this approach here doesn't look easy, with $D_{24}$ involved.
Could anyone suggest another way of solving this ?
abstract-algebra group-theory dihedral-groups
edited Nov 27 '18 at 21:35
davidlowryduda♦
74.4k7117251
asked Sep 16 '15 at 11:07
User9523
9701823
|
show 10 more comments
We need to prove that $D_{3} oplus D_{4}$ is not isomorphic to $D_{24}$ .
The way in which I approach such type of questions is to count the number of elements of order $x$ in one group and then in the other group, and then conclude that they aren't equal and hence there can't be any isomorphism between them.
But this approach here doesn't look easy, with $D_{24}$ involved.
Could anyone suggest another way of solving this ?
abstract-algebra group-theory dihedral-groups
edited Nov 27 '18 at 21:35
davidlowryduda♦
74.4k7117251
asked Sep 16 '15 at 11:07
User9523
9701823
3
Well, you need to start with the right order. $x = 24$ is a good choice to start with.
– Daniel Fischer♦
Sep 16 '15 at 11:11
1
Yes, $a^{12} =1 $ for all $ain D_3 times D_4$.
– peter a g
Sep 16 '15 at 11:57
What I found is that$D_{3} oplus D_{4}$ has no element of order $24$.. And I have no idea about $D_{24}$ , could you please help me further ? @DanielFischer
– User9523
Sep 16 '15 at 13:56
2
You don't need to know how many elements of order $24$ there are in $D_{24}$. It suffices if you know that the number is not $0$. But in $D_n$, there always is an element of order $n$, so done.
– Daniel Fischer♦
Sep 16 '15 at 13:59
Got it.. Thanks :) @DanielFischer
– User9523
Sep 16 '15 at 15:57
|
show 10 more comments
We need to prove that $D_{3} oplus D_{4}$ is not isomorphic to $D_{24}$ .
The way in which I approach such type of questions is to count the number of elements of order $x$ in one group and then in the other group, and then conclude that they aren't equal and hence there can't be any isomorphism between them.
But this approach here doesn't look easy, with $D_{24}$ involved.
Could anyone suggest another way of solving this ?
abstract-algebra group-theory dihedral-groups
edited Nov 27 '18 at 21:35
davidlowryduda♦
74.4k7117251
asked Sep 16 '15 at 11:07
User9523
9701823
We need to prove that $D_{3} oplus D_{4}$ is not isomorphic to $D_{24}$ .
The way in which I approach such type of questions is to count the number of elements of order $x$ in one group and then in the other group, and then conclude that they aren't equal and hence there can't be any isomorphism between them.
But this approach here doesn't look easy, with $D_{24}$ involved.
Could anyone suggest another way of solving this ?
abstract-algebra group-theory dihedral-groups
abstract-algebra group-theory dihedral-groups
edited Nov 27 '18 at 21:35
davidlowryduda♦
74.4k7117251
asked Sep 16 '15 at 11:07
User9523
9701823
edited Nov 27 '18 at 21:35
davidlowryduda♦
74.4k7117251
asked Sep 16 '15 at 11:07
User9523
9701823
edited Nov 27 '18 at 21:35
davidlowryduda♦
74.4k7117251
edited Nov 27 '18 at 21:35
davidlowryduda♦
74.4k7117251
edited Nov 27 '18 at 21:35
davidlowryduda♦
74.4k7117251
74.4k7117251
asked Sep 16 '15 at 11:07
User9523
9701823
asked Sep 16 '15 at 11:07
User9523
9701823
asked Sep 16 '15 at 11:07
User9523
9701823
9701823
3
Well, you need to start with the right order. $x = 24$ is a good choice to start with.
– Daniel Fischer♦
Sep 16 '15 at 11:11
1
Yes, $a^{12} =1 $ for all $ain D_3 times D_4$.
– peter a g
Sep 16 '15 at 11:57
What I found is that$D_{3} oplus D_{4}$ has no element of order $24$.. And I have no idea about $D_{24}$ , could you please help me further ? @DanielFischer
– User9523
Sep 16 '15 at 13:56
2
You don't need to know how many elements of order $24$ there are in $D_{24}$. It suffices if you know that the number is not $0$. But in $D_n$, there always is an element of order $n$, so done.
– Daniel Fischer♦
Sep 16 '15 at 13:59
Got it.. Thanks :) @DanielFischer
– User9523
Sep 16 '15 at 15:57
|
show 10 more comments
3
Well, you need to start with the right order. $x = 24$ is a good choice to start with.
– Daniel Fischer♦
Sep 16 '15 at 11:11
1
Yes, $a^{12} =1 $ for all $ain D_3 times D_4$.
– peter a g
Sep 16 '15 at 11:57
What I found is that$D_{3} oplus D_{4}$ has no element of order $24$.. And I have no idea about $D_{24}$ , could you please help me further ? @DanielFischer
– User9523
Sep 16 '15 at 13:56
2
You don't need to know how many elements of order $24$ there are in $D_{24}$. It suffices if you know that the number is not $0$. But in $D_n$, there always is an element of order $n$, so done.
– Daniel Fischer♦
Sep 16 '15 at 13:59
Got it.. Thanks :) @DanielFischer
– User9523
Sep 16 '15 at 15:57
3
3
Well, you need to start with the right order. $x = 24$ is a good choice to start with.
– Daniel Fischer♦
Sep 16 '15 at 11:11
Well, you need to start with the right order. $x = 24$ is a good choice to start with.
– Daniel Fischer♦
Sep 16 '15 at 11:11
1
1
Yes, $a^{12} =1 $ for all $ain D_3 times D_4$.
– peter a g
Sep 16 '15 at 11:57
Yes, $a^{12} =1 $ for all $ain D_3 times D_4$.
– peter a g
Sep 16 '15 at 11:57
What I found is that$D_{3} oplus D_{4}$ has no element of order $24$.. And I have no idea about $D_{24}$ , could you please help me further ? @DanielFischer
– User9523
Sep 16 '15 at 13:56
What I found is that$D_{3} oplus D_{4}$ has no element of order $24$.. And I have no idea about $D_{24}$ , could you please help me further ? @DanielFischer
– User9523
Sep 16 '15 at 13:56
2
2
You don't need to know how many elements of order $24$ there are in $D_{24}$. It suffices if you know that the number is not $0$. But in $D_n$, there always is an element of order $n$, so done.
– Daniel Fischer♦
Sep 16 '15 at 13:59
You don't need to know how many elements of order $24$ there are in $D_{24}$. It suffices if you know that the number is not $0$. But in $D_n$, there always is an element of order $n$, so done.
– Daniel Fischer♦
Sep 16 '15 at 13:59
Got it.. Thanks :) @DanielFischer
– User9523
Sep 16 '15 at 15:57
Got it.. Thanks :) @DanielFischer
– User9523
Sep 16 '15 at 15:57
|
show 10 more comments
3 Answers
3
active
oldest
votes
Here is a stronger result (related to this thread). I claim that the dihedral group $D_n$ is decomposable if and only if $nequiv 2pmod{4}$. (In particular, when $n=24$, we have $nnotequiv 2pmod{4}$, so $D_n=D_{24}$ is indecomposable, so you can't write $D_{24}=D_3times D_4$.) The case $nin{1,2}$ is easy. We now assume that $ngeq 3$.
If $D_n=Atimes B$ for some non-trivial subgroups $A$ and $B$, then $A$ and $B$ are normal subgroups of $D_n$, and $Acap B={1}$. Write $D_n=langle R,Frangle$, where $R^n=F^2=1$ and $(RF)^2=1$ as given here. Then, the normal subgroups of $D_n$ are
$n$ odd: $langle R^drangle$ where $d$ is a divisor of $n$;
$n$ even: $langle R^drangle$ where $d$ is a divisor of $n$, and two more $langle R^2,Frangle$ and $langle R^2,RFrangle$.
So, if $n$ is odd, then both $A$ and $B$ are subgroups of the cyclic subgroup $langle Rrangle$, which means $A$ and $B$ are abelian, and so $D_n=Atimes B$ is abelian, which is a contradiciton.
If $ngeq 4$ is even, then exactly one of $A$ and $B$ takes the form $langle R^2,Frangle$ or $langle R^2,RFrangle$. WLOG, $A$ is of such a form. Then, $Acong D_{n/2}$. So, $B$ must be isomorphic to the cyclic group $C_2$ of order $2$.
If $nequiv 2pmod{4}$, then we can take $A=langle R^2,Frangle$ and $B=langle R^{n/2}rangle$ to see that $$D_n=Atimes Bcong D_{n/2}times C_2.$$ In fact, there are exactly two ways to decompose $D_n$ into a direct product of non-trivial subgroups:
$$D_n=langle R^2,Frangle times langle R^{n/2}rangle$$
and
$$D_n=langle R^2,RFrangle times langle R^{n/2}rangle.$$
If $4mid n$, then all elements of order $2$ of $D_n$ are $R^{n/2}$ and $R^kF$. But the subgroup generated by $R^kF$ is not normal, and $R^{n/2} in A$ because $frac{n}{2}$ is even. Therefore, $A$ and $B$ cannot exist. Hence, $D_n$ is indecomposable when $4mid n$.
answered Nov 20 '18 at 15:12
Zvi
4,770430
add a comment |
Hint The group $D_{24}$ (usually denoted $D_{48}$) has an element of order $24$. Does $D_3 times D_4$?
answered Nov 20 '18 at 14:52
Travis
59.7k767146
2
Maximum possible order of element in $D_{3} = 3$ and in $D_{4}=4$. So maximum possible order of element in $D_{3}oplus D_{4} = lcm(3,4)= 12$. Will this work?
– So Lo
Nov 20 '18 at 14:54
Yes!$!!!!!$
– Travis
Nov 20 '18 at 15:06
add a comment |
HINT: Look at the commutator subgroups of elements of order three in both groups.
answered Nov 20 '18 at 14:48
Josué Tonelli-Cueto
3,6721027
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%2f1437850%2fd-3-oplus-d-4-not-isomorphic-to-d-24%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here is a stronger result (related to this thread). I claim that the dihedral group $D_n$ is decomposable if and only if $nequiv 2pmod{4}$. (In particular, when $n=24$, we have $nnotequiv 2pmod{4}$, so $D_n=D_{24}$ is indecomposable, so you can't write $D_{24}=D_3times D_4$.) The case $nin{1,2}$ is easy. We now assume that $ngeq 3$.
If $D_n=Atimes B$ for some non-trivial subgroups $A$ and $B$, then $A$ and $B$ are normal subgroups of $D_n$, and $Acap B={1}$. Write $D_n=langle R,Frangle$, where $R^n=F^2=1$ and $(RF)^2=1$ as given here. Then, the normal subgroups of $D_n$ are
$n$ odd: $langle R^drangle$ where $d$ is a divisor of $n$;
$n$ even: $langle R^drangle$ where $d$ is a divisor of $n$, and two more $langle R^2,Frangle$ and $langle R^2,RFrangle$.
So, if $n$ is odd, then both $A$ and $B$ are subgroups of the cyclic subgroup $langle Rrangle$, which means $A$ and $B$ are abelian, and so $D_n=Atimes B$ is abelian, which is a contradiciton.
If $ngeq 4$ is even, then exactly one of $A$ and $B$ takes the form $langle R^2,Frangle$ or $langle R^2,RFrangle$. WLOG, $A$ is of such a form. Then, $Acong D_{n/2}$. So, $B$ must be isomorphic to the cyclic group $C_2$ of order $2$.
If $nequiv 2pmod{4}$, then we can take $A=langle R^2,Frangle$ and $B=langle R^{n/2}rangle$ to see that $$D_n=Atimes Bcong D_{n/2}times C_2.$$ In fact, there are exactly two ways to decompose $D_n$ into a direct product of non-trivial subgroups:
$$D_n=langle R^2,Frangle times langle R^{n/2}rangle$$
and
$$D_n=langle R^2,RFrangle times langle R^{n/2}rangle.$$
If $4mid n$, then all elements of order $2$ of $D_n$ are $R^{n/2}$ and $R^kF$. But the subgroup generated by $R^kF$ is not normal, and $R^{n/2} in A$ because $frac{n}{2}$ is even. Therefore, $A$ and $B$ cannot exist. Hence, $D_n$ is indecomposable when $4mid n$.
answered Nov 20 '18 at 15:12
Zvi
4,770430
add a comment |
Here is a stronger result (related to this thread). I claim that the dihedral group $D_n$ is decomposable if and only if $nequiv 2pmod{4}$. (In particular, when $n=24$, we have $nnotequiv 2pmod{4}$, so $D_n=D_{24}$ is indecomposable, so you can't write $D_{24}=D_3times D_4$.) The case $nin{1,2}$ is easy. We now assume that $ngeq 3$.
If $D_n=Atimes B$ for some non-trivial subgroups $A$ and $B$, then $A$ and $B$ are normal subgroups of $D_n$, and $Acap B={1}$. Write $D_n=langle R,Frangle$, where $R^n=F^2=1$ and $(RF)^2=1$ as given here. Then, the normal subgroups of $D_n$ are
$n$ odd: $langle R^drangle$ where $d$ is a divisor of $n$;
$n$ even: $langle R^drangle$ where $d$ is a divisor of $n$, and two more $langle R^2,Frangle$ and $langle R^2,RFrangle$.
So, if $n$ is odd, then both $A$ and $B$ are subgroups of the cyclic subgroup $langle Rrangle$, which means $A$ and $B$ are abelian, and so $D_n=Atimes B$ is abelian, which is a contradiciton.
If $ngeq 4$ is even, then exactly one of $A$ and $B$ takes the form $langle R^2,Frangle$ or $langle R^2,RFrangle$. WLOG, $A$ is of such a form. Then, $Acong D_{n/2}$. So, $B$ must be isomorphic to the cyclic group $C_2$ of order $2$.
If $nequiv 2pmod{4}$, then we can take $A=langle R^2,Frangle$ and $B=langle R^{n/2}rangle$ to see that $$D_n=Atimes Bcong D_{n/2}times C_2.$$ In fact, there are exactly two ways to decompose $D_n$ into a direct product of non-trivial subgroups:
$$D_n=langle R^2,Frangle times langle R^{n/2}rangle$$
and
$$D_n=langle R^2,RFrangle times langle R^{n/2}rangle.$$
If $4mid n$, then all elements of order $2$ of $D_n$ are $R^{n/2}$ and $R^kF$. But the subgroup generated by $R^kF$ is not normal, and $R^{n/2} in A$ because $frac{n}{2}$ is even. Therefore, $A$ and $B$ cannot exist. Hence, $D_n$ is indecomposable when $4mid n$.
answered Nov 20 '18 at 15:12
Zvi
4,770430
add a comment |
Here is a stronger result (related to this thread). I claim that the dihedral group $D_n$ is decomposable if and only if $nequiv 2pmod{4}$. (In particular, when $n=24$, we have $nnotequiv 2pmod{4}$, so $D_n=D_{24}$ is indecomposable, so you can't write $D_{24}=D_3times D_4$.) The case $nin{1,2}$ is easy. We now assume that $ngeq 3$.
If $D_n=Atimes B$ for some non-trivial subgroups $A$ and $B$, then $A$ and $B$ are normal subgroups of $D_n$, and $Acap B={1}$. Write $D_n=langle R,Frangle$, where $R^n=F^2=1$ and $(RF)^2=1$ as given here. Then, the normal subgroups of $D_n$ are
$n$ odd: $langle R^drangle$ where $d$ is a divisor of $n$;
$n$ even: $langle R^drangle$ where $d$ is a divisor of $n$, and two more $langle R^2,Frangle$ and $langle R^2,RFrangle$.
So, if $n$ is odd, then both $A$ and $B$ are subgroups of the cyclic subgroup $langle Rrangle$, which means $A$ and $B$ are abelian, and so $D_n=Atimes B$ is abelian, which is a contradiciton.
If $ngeq 4$ is even, then exactly one of $A$ and $B$ takes the form $langle R^2,Frangle$ or $langle R^2,RFrangle$. WLOG, $A$ is of such a form. Then, $Acong D_{n/2}$. So, $B$ must be isomorphic to the cyclic group $C_2$ of order $2$.
If $nequiv 2pmod{4}$, then we can take $A=langle R^2,Frangle$ and $B=langle R^{n/2}rangle$ to see that $$D_n=Atimes Bcong D_{n/2}times C_2.$$ In fact, there are exactly two ways to decompose $D_n$ into a direct product of non-trivial subgroups:
$$D_n=langle R^2,Frangle times langle R^{n/2}rangle$$
and
$$D_n=langle R^2,RFrangle times langle R^{n/2}rangle.$$
If $4mid n$, then all elements of order $2$ of $D_n$ are $R^{n/2}$ and $R^kF$. But the subgroup generated by $R^kF$ is not normal, and $R^{n/2} in A$ because $frac{n}{2}$ is even. Therefore, $A$ and $B$ cannot exist. Hence, $D_n$ is indecomposable when $4mid n$.
answered Nov 20 '18 at 15:12
Zvi
4,770430
Here is a stronger result (related to this thread). I claim that the dihedral group $D_n$ is decomposable if and only if $nequiv 2pmod{4}$. (In particular, when $n=24$, we have $nnotequiv 2pmod{4}$, so $D_n=D_{24}$ is indecomposable, so you can't write $D_{24}=D_3times D_4$.) The case $nin{1,2}$ is easy. We now assume that $ngeq 3$.
If $D_n=Atimes B$ for some non-trivial subgroups $A$ and $B$, then $A$ and $B$ are normal subgroups of $D_n$, and $Acap B={1}$. Write $D_n=langle R,Frangle$, where $R^n=F^2=1$ and $(RF)^2=1$ as given here. Then, the normal subgroups of $D_n$ are
$n$ odd: $langle R^drangle$ where $d$ is a divisor of $n$;
$n$ even: $langle R^drangle$ where $d$ is a divisor of $n$, and two more $langle R^2,Frangle$ and $langle R^2,RFrangle$.
So, if $n$ is odd, then both $A$ and $B$ are subgroups of the cyclic subgroup $langle Rrangle$, which means $A$ and $B$ are abelian, and so $D_n=Atimes B$ is abelian, which is a contradiciton.
If $ngeq 4$ is even, then exactly one of $A$ and $B$ takes the form $langle R^2,Frangle$ or $langle R^2,RFrangle$. WLOG, $A$ is of such a form. Then, $Acong D_{n/2}$. So, $B$ must be isomorphic to the cyclic group $C_2$ of order $2$.
If $nequiv 2pmod{4}$, then we can take $A=langle R^2,Frangle$ and $B=langle R^{n/2}rangle$ to see that $$D_n=Atimes Bcong D_{n/2}times C_2.$$ In fact, there are exactly two ways to decompose $D_n$ into a direct product of non-trivial subgroups:
$$D_n=langle R^2,Frangle times langle R^{n/2}rangle$$
and
$$D_n=langle R^2,RFrangle times langle R^{n/2}rangle.$$
If $4mid n$, then all elements of order $2$ of $D_n$ are $R^{n/2}$ and $R^kF$. But the subgroup generated by $R^kF$ is not normal, and $R^{n/2} in A$ because $frac{n}{2}$ is even. Therefore, $A$ and $B$ cannot exist. Hence, $D_n$ is indecomposable when $4mid n$.
answered Nov 20 '18 at 15:12
Zvi
4,770430
edited Nov 20 '18 at 19:35
answered Nov 20 '18 at 15:12
Zvi
4,770430
answered Nov 20 '18 at 15:12
Zvi
4,770430
answered Nov 20 '18 at 15:12
Zvi
4,770430
4,770430
add a comment |
add a comment |
Hint The group $D_{24}$ (usually denoted $D_{48}$) has an element of order $24$. Does $D_3 times D_4$?
answered Nov 20 '18 at 14:52
Travis
59.7k767146
2
Maximum possible order of element in $D_{3} = 3$ and in $D_{4}=4$. So maximum possible order of element in $D_{3}oplus D_{4} = lcm(3,4)= 12$. Will this work?
– So Lo
Nov 20 '18 at 14:54
Yes!$!!!!!$
– Travis
Nov 20 '18 at 15:06
add a comment |
Hint The group $D_{24}$ (usually denoted $D_{48}$) has an element of order $24$. Does $D_3 times D_4$?
answered Nov 20 '18 at 14:52
Travis
59.7k767146
2
Maximum possible order of element in $D_{3} = 3$ and in $D_{4}=4$. So maximum possible order of element in $D_{3}oplus D_{4} = lcm(3,4)= 12$. Will this work?
– So Lo
Nov 20 '18 at 14:54
Yes!$!!!!!$
– Travis
Nov 20 '18 at 15:06
add a comment |
Hint The group $D_{24}$ (usually denoted $D_{48}$) has an element of order $24$. Does $D_3 times D_4$?
answered Nov 20 '18 at 14:52
Travis
59.7k767146
Hint The group $D_{24}$ (usually denoted $D_{48}$) has an element of order $24$. Does $D_3 times D_4$?
answered Nov 20 '18 at 14:52
Travis
59.7k767146
answered Nov 20 '18 at 14:52
Travis
59.7k767146
answered Nov 20 '18 at 14:52
Travis
59.7k767146
answered Nov 20 '18 at 14:52
Travis
59.7k767146
59.7k767146
2
Maximum possible order of element in $D_{3} = 3$ and in $D_{4}=4$. So maximum possible order of element in $D_{3}oplus D_{4} = lcm(3,4)= 12$. Will this work?
– So Lo
Nov 20 '18 at 14:54
Yes!$!!!!!$
– Travis
Nov 20 '18 at 15:06
add a comment |
2
Maximum possible order of element in $D_{3} = 3$ and in $D_{4}=4$. So maximum possible order of element in $D_{3}oplus D_{4} = lcm(3,4)= 12$. Will this work?
– So Lo
Nov 20 '18 at 14:54
Yes!$!!!!!$
– Travis
Nov 20 '18 at 15:06
2
2
Maximum possible order of element in $D_{3} = 3$ and in $D_{4}=4$. So maximum possible order of element in $D_{3}oplus D_{4} = lcm(3,4)= 12$. Will this work?
– So Lo
Nov 20 '18 at 14:54
Maximum possible order of element in $D_{3} = 3$ and in $D_{4}=4$. So maximum possible order of element in $D_{3}oplus D_{4} = lcm(3,4)= 12$. Will this work?
– So Lo
Nov 20 '18 at 14:54
Yes!$!!!!!$
– Travis
Nov 20 '18 at 15:06
Yes!$!!!!!$
– Travis
Nov 20 '18 at 15:06
add a comment |
HINT: Look at the commutator subgroups of elements of order three in both groups.
answered Nov 20 '18 at 14:48
Josué Tonelli-Cueto
3,6721027
add a comment |
HINT: Look at the commutator subgroups of elements of order three in both groups.
answered Nov 20 '18 at 14:48
Josué Tonelli-Cueto
3,6721027
add a comment |
HINT: Look at the commutator subgroups of elements of order three in both groups.
answered Nov 20 '18 at 14:48
Josué Tonelli-Cueto
3,6721027
HINT: Look at the commutator subgroups of elements of order three in both groups.
answered Nov 20 '18 at 14:48
Josué Tonelli-Cueto
3,6721027
answered Nov 20 '18 at 14:48
Josué Tonelli-Cueto
3,6721027
answered Nov 20 '18 at 14:48
Josué Tonelli-Cueto
3,6721027
answered Nov 20 '18 at 14:48
Josué Tonelli-Cueto
3,6721027
3,6721027
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f1437850%2fd-3-oplus-d-4-not-isomorphic-to-d-24%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
3
Well, you need to start with the right order. $x = 24$ is a good choice to start with.
– Daniel Fischer♦
Sep 16 '15 at 11:11
1
Yes, $a^{12} =1 $ for all $ain D_3 times D_4$.
– peter a g
Sep 16 '15 at 11:57
What I found is that$D_{3} oplus D_{4}$ has no element of order $24$.. And I have no idea about $D_{24}$ , could you please help me further ? @DanielFischer
– User9523
Sep 16 '15 at 13:56
2
You don't need to know how many elements of order $24$ there are in $D_{24}$. It suffices if you know that the number is not $0$. But in $D_n$, there always is an element of order $n$, so done.
– Daniel Fischer♦
Sep 16 '15 at 13:59
Got it.. Thanks :) @DanielFischer
– User9523
Sep 16 '15 at 15:57