2

In $S_4$, what does the expression for the cyclic group $\langle (13),(1234)\rangle$ mean?

I apologize if this is too basic, but I haven't come across such an expression anywhere in my book. Also, not much success on the web regarding what does $\langle (13),(1234)\rangle$ actually mean? How is it evaluated?

Thank you for your help.

MathMan
  • 8,974
  • 7
  • 70
  • 135
  • 2
    It is the subgroup generated by $(13)$ and $(1234)$. You can get the elements explicitly by multiplying $(13)$ by itself and $(1234)$ until you no longer get any new elements, and then you multiply every combination of those elements. In words, it is the smallest subgroup of $S_4$ containing those elements. – dannum Aug 25 '14 at 11:58
  • 1
    Also, here's more of an explanation. See page 4 http://www.math.uiuc.edu/~kapovich/427-13/sgps.pdf – dannum Aug 25 '14 at 12:01

2 Answers2

4

This is not "cyclic", that would be the case if it is only one element of $S_4$. But this is a generalization - the group generated by two elements namely $(13)$ and $(1234)$. So write down all elements that can be made with powers and products of these elements, for example $(13)\cdot(1234)\cdot(1234)\cdot (13)\cdot (1234)$ is an element. Can you work out which subgroup of $S_4$ the group $\langle(13),(1234)\rangle$ represents?

Nicky Hekster
  • 49,281
  • ohk .. so this group should comprise the following elements : ${ 1,(13),(1234),(1234)^2,(1234)^3,(13)(1234),(13)(1234)^2,(13)(1234)^3}$ – MathMan Aug 25 '14 at 12:09
  • Am I correct .. ? – MathMan Aug 25 '14 at 12:11
  • 1
    And to be completely clear, "powers" includes negative powers too. Another way to describe $\langle a,b\rangle$ is "all finite products you can make only using $a,b,a^{-1}$ and $b^{-1}$. For finite groups one can get away with just "powers," but when we get to infinite groups we'll be glad we mentioned this :) – rschwieb Aug 25 '14 at 12:53
  • @nicky and @ rschwieb Thank you for your answers :-) – MathMan Aug 25 '14 at 14:12
  • 1
    Very good remarks of all, and yes @VHP you were correct in identifying the group - it is isomorphic to $D_4$. – Nicky Hekster Aug 25 '14 at 21:04
  • The way isomorphism to $D_4$ was or should be determined was through examining the orders of elements on both sides right? – MathMan Aug 25 '14 at 21:31
  • Well, be careful, in general, if two groups have the same number of elements of a specific order, then they DO NOT have to be isomorphic. This has been discussed on this forum before - have to look up the reference. Most of the time the presentation of $D_4$ is $\langle a,b : a^4=b^2=1, bab=a^{-1}\rangle$. From this you can construct an isomorphism mapping $(1234)$ to $a$ and $(13)$ to $b$. – Nicky Hekster Aug 25 '14 at 21:41
  • See http://math.stackexchange.com/questions/693163/groups-with-same-number-of-elements-of-each-order for my remark above. – Nicky Hekster Aug 25 '14 at 21:45
2

$\langle(13),(1234)\rangle$ is NOT a cyclic group! A cyclic group is a group generated by a single element, the one you wrote is a subgroup of $S_4$ generated by TWO elements, $(13)$ and $(1234)$.

In general given a group $G$, taken $g_1,\dots,g_n\in G$, the expression $\langle g_1,\dots,g_n\rangle$ denotes the smallest subgroup of $G$ which contains the $g_i$'s which is the intersection of all the subgroups of $G$ containing all the $g_i$'s.

In your case, you can write down explicitly the elements of the subgroup $\langle(13),(1234)\rangle$: it's formed by all the product of integers power of the elements $(13)$ and $(1234)$. Hence a typical element of $\langle(13),(1234)\rangle$ is $(13)^{k_1}(1234)^{k_2}(13)^{k_3}\dots$ with $k_i\in\mathbb Z$. Being $(13)$ an element of order $2$ you can wlog think the element of your group as $(13)(1234)^{h_1}(13)(1234)^{h_2}\dots$ and $(1234)^{h_1}(13)(1234)^{h_2}(13)\dots$ with $h_i\in\mathbb Z$.

Joe
  • 11,745
  • I got this group as : ${ 1,(13),(1234),(1234)^2,(1234)^3,(13)(1234),(13)(1234)^2,(13)(1234)^3}$ .. I just saw on another webpage that this is isomorphic to $D_4$ . In general, how should we find out $D_n$ is isomorphic to which subgroup of $S_4$ ? – MathMan Aug 25 '14 at 12:18
  • 1
    $D_n$ has order $2n$ hence being $|S_4|=24$ and being its subgroups of order dividing $24$, the only possibilities for $D_n$ to be isomorphic to a subgroup of $S_4$ is for (assumed $n>4$) $n=6$. But $|D_6|=12$ and the only subgroup of $S_4$ of such an order is $A_4$. Check if $D_6$ could be isomorphic to $A_4$ (look for example at the order of certain elements). – Joe Aug 25 '14 at 12:33
  • Thank you for your answer :-) – MathMan Aug 25 '14 at 14:11
  • You're welcome ;-) – Joe Aug 25 '14 at 14:34