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Let $G = \langle(12)(34), (15)\rangle$ be a subgroup of $S_5$.

Then I need to show that $G$ has order $12$ and has a non-trivial centre.

I have found thse elements- $$I,(12)(34), (15), (12)(34)(15), (15)(12)(34).$$

If I just keep computing compositions, then the whole process is becoming really cumbersome.

Please, help!

Shaun
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Shivani
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3 Answers3

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This $G$ is a group generated by two involutions. So it is a dihedral group $D_n$ for some $n$. This $n$ is the order of the product of the involutions $(15)(12)(34)=(152)(34)$ which is $6$, so your group $G$ is isomorphic to $D_6$ and has 12 elements.

markvs
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    The product is $(1,2,5)(3,4)$. – David A. Craven Jul 17 '20 at 20:21
  • You are composing left-to-right, correct? – Arturo Magidin Jul 17 '20 at 20:33
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    The nice thing about standards is that there are so many to choose from... Both Hungerford and Herstein, to name but two, multiply permutations right-to-left, like function compositions, so that $(15)(12)=(125)$, not $(152)$. – Arturo Magidin Jul 18 '20 at 00:31
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    @ArturoMagidin my comment was because the original version of the answer said that the two involutions commuted, not that I was arguing the toss between LtR and RtL. (I'm a finite group theorist, and most of those write LtR.) – David A. Craven Jul 18 '20 at 07:03
  • @DavidCraven: I work in groups too, and have spent a fair amount of the previous years on finite $p$-groups. I think that, like the issue of “what does extension of $A$ by $B$ mean?”, it’s even more area-specific. I expect that in areas where you do a lot of actions and representations, and in which right actions are prefered, left-to-right is the standard; while in other areas where there is no strong preference for right actions, or preference for left, right-to-left is common. – Arturo Magidin Jul 18 '20 at 20:01
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    @ArturoMagidin Oh, the extension of $A$ by $B$ thing is really annoying. Certainly the finite simple groups people write almost exclusively LtR. Some of the older $p$-groups people I think wrote LtR, but a lot write RtL nowadays. A lot of representation theory people, particular modular representation theorists, come from a background that wasn't hardcore finite group theory, and so write RtL. Magma and GAP do LtR for permutations, and use right modules, etc. It's all a complete mess. As I wrote in my recent book, if you violently prefer left actions hold the book in a mirror, or use a pencil. – David A. Craven Jul 18 '20 at 20:25
  • Cont.: at least, that's my impression, based on nothing more than anecdotes, hearsay and prejudice. – David A. Craven Jul 18 '20 at 20:34
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Let $a = (12)(34)$ and $b = (15)$. Clearly $a^2 = b^2 = e$, where $e$ is the identity.

Define $u = ab$. It is easy to check that $u$ has order six, and also that $auau = e$. I can add detail if you need me to, but this seems like a good exercise to me.

The dihedral group of order $12$ (the symmetries of a regular hexagon) has presentation $$ D_{12} = \langle \sigma, \tau \mid \sigma^6 = \tau^2 = \sigma \tau\sigma\tau = e \rangle $$

We have verified that $a, u$ satisfy the defining relations of $D_{12}$, which means that the group generated by $a$ and $u$ has size at most $12$ (intuitively, this is because any combination of them that equals the identity in $D_{12}$ will also equal the identity in $G$, so there can be at most twelve distinct combinations. This follows very easily from some slightly more advanced theory but I'll try to avoid that).

Since $b = a^{-1}u$, the group generated by $a$ and $u$ is actually, $G$, so $G$ has size at most twelve. But $u$ is an element of order six, so the order of $G$ is a multiple of six. The cyclic group of order six has only one element of order six, but $a$ and $b$ are distinct elements of order $2$, which means that $G \neq \langle u \rangle \cong C_6$, and hence $\lvert G \rvert > 6$, so $\lvert G \rvert = 12$.

Sebastian Monnet
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One way is to use GAP, like so:

gap> S:=SymmetricGroup(5);
Sym( [ 1 .. 5 ] )
gap> G:=Subgroup(S, [(1,2)*(3,4), (1,5)]);
Group([ (1,2)(3,4), (1,5) ])
gap> Size(G);
12
gap> Centre(G);
Group([ (3,4) ])
gap> Size(last);
2
gap>
Shaun
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    G:=Group( (1,2)(3,4),(1,5) ); also works (for permutation groups, GAP assumes that they all naturally embed in each others, if points match). Also, there is no need for a star in (1,2)(3,4) if cycles are independent. – Olexandr Konovalov Jul 18 '20 at 16:18
  • Thank you for the tip, @AlexanderKonovalov :) – Shaun Jul 18 '20 at 20:18