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Show that the presentation $G=\langle a,b,c\mid a^2 = b^2 = c^3 = 1, ab = ba, cac^{-1} = b, cbc^{-1} =ab\rangle$ defines a group of order $12$.

I tried to let $d=ab\Rightarrow G=\langle d,c\mid d^2 =c^3 = 1, c^2d=dcdc\rangle$. But I don't know how to find the order of the new presentation. I mean I am not sure how the elements of new $G$ look like. (For sure not on the form $c^id^j$ and $d^kc^l$ otherwise $|G|\leq 5$).

Is it good step to reduce the number of generators or not necessary?

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6 Answers 6

Direct proof.

$N:=\langle a,b \rangle$ is clearly a normal subgroup of $G$ with $G/N = \langle c : c^3 = 1 \rangle = C_3$, and $N$ is a quotient of the Klein four-group $V_4 = \langle a,b : a^2=b^2=1, ab=ba \rangle$. Therefore $|G|$ divides $12$. Now one checks that the permutations $a=(12)(34)$, $b = (14)(23)$, $c = (123) \in A_4$ satisfy the relations which define $G$, and that they generate $A_4$. Hence we get a surjective homomorphism $G \to A_4$. Since $|G|$ divides $12$, this shows that $G \cong A_4$ is an isomorphism and $|G|=12$.

A more conceptual proof using semidirect products.

The automorphism group of the Klein four-group $V_4$ is $S_3$ (since you can permute $a,b,ab$ freely, you can also see this from linear algebra applied to $V_4 \cong \mathbb{F}_2^2$). In particular there is an automorphism $c$ of order $3$, namely the one mapping $a \mapsto b, b \mapsto ab, ab \mapsto a$. This means that the cyclic group $C_3 = \langle c : c^3 = 1 \rangle$ of order $3$ acts on this group. The corresponding semidirect product $V_4 \rtimes C_3$ has the desired presentation $\langle a,b,c : a^2=b^2=c^3=1, ab=ba, c a c^{-1} = b, c b c^{-1} = ab \rangle$. See MO/96078 for the universal property and the resulting group presentation of a semidirect product. But the usual construction of the semidirect product $N \rtimes H$ shows that $|N \rtimes H|= |N| |H|$. In particular, $G \cong V_4 \rtimes C_3$ has order $12$.

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Before editing was easier to follow –  Ronald Aug 17 '13 at 19:02
    
this is very clear and useful .. thanks a lot –  Ronald Aug 17 '13 at 19:34
    
But since $N$ is normal why can't we directly use $|G|=|N||\frac{G}{N}|$ hence $|G|=12$? –  Ronald Aug 17 '13 at 19:55
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A priori $N$ is only a quotient of $V_4$ (since $a,b \in N$ satisfy $a^2=b^2=1$ and $ab=ba$). One has to prove that $N=V_4$ (equivalently that $|N|=4$). For example, one has to check that $a \neq 1$ and that $b \neq 1$. This is only possible by giving a specific group where the relations for $a,b,c$ hold, but $a=b^2$ does not hold. In fact, there is no general algorithm for groups which checks whether a relation follows from other relations (as compared to the case of monoids). –  Martin Brandenburg Aug 17 '13 at 20:55

Although my approach is not the way you expected, it is a nice way for your group. This way is called Coset enumeration or Todd-Coxeter Algorithm. You have see $12$ rows completed as follows. In fact, I found $[G:\langle e\rangle]=12$:

enter image description here

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It looks simple,, I will learn this way .. thank you –  Ronald Aug 17 '13 at 19:35
    
It's quicker to do the coset enumeration using the subgroup $H:=\langle c \rangle$ and prove that $|G:H|=4$. –  Derek Holt Aug 17 '13 at 21:47
    
Nice work, Babak! +1 –  amWhy Aug 18 '13 at 0:18
    
@Babak Great! +1 –  Mahdi Khosravi Aug 18 '13 at 5:41
    
@MahdiKhosravi: G.M. Mahdiand Thanks . I think this the first time someone did this way for solving this special question. It took time to be completed but it was enjoyable reaching the target. Moreover, what Derek commented means a lot for me and show me he had a look at mine here. I am glad of that. :-) –  Babak S. Aug 18 '13 at 5:45

Consider $D := \langle a, b\rangle$ and $C = \langle c \rangle$. Note that

$$ab = ba \implies D = \langle a \rangle \langle b \rangle$$

so $|D| \leq 4$. Further, the last two relations tell you that $c \in N_{G}(D)$ (the normalizer in $G$ of $D$), so $C \leq N_{G}(D)$. It follows that $DC \leq G$, but $DC$ contains every generator of $C$; therefore, $DC = G$. We then have

$$|G| = |DC| = \frac{|D||C|}{|D \cap C|} \leq \frac{|D||C|}{1} \leq \frac{4 \cdot 3}{1} = 12$$

It just remains to find a group of order $12$ satisfying the given relations - so how many groups of order $12$ do you know?

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It is $A_4$ thanks:-) –  Ronald Aug 17 '13 at 20:10
    
But I need some explanation where did you use that $C \leq N_{G}(D)$ please –  Ronald Aug 17 '13 at 20:21
    
@Danial In order to conclude that $CD \leq G$. –  user61527 Aug 17 '13 at 20:28
    
Isn't $D$ is normal, and so $DH\leq G$ for any $H\leq G$? –  Ronald Aug 17 '13 at 20:31
    
Also, Is any subgroup of a normalizer is normal in $G$? Or is it a fact that if $A\leq N_G(B)$ then $AB\leq G$? –  Ronald Aug 17 '13 at 20:33

If you note in the original presentation that the exponents of $a,b,c$ are $2,2,3$ whose product is $12$ a natural strategy would be to try to put a general element in the form $a^pb^qc^r$ and then show that these elements are distinct.

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Yes, then is the original presentation easier to find the order? –  Ronald Aug 17 '13 at 18:47
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It depends what you mean by "easier" - the first presentation shows that the group generated by $a, b$ is normal and reveals a lot of structure. I almost deleted my answer when I saw Martin Brandenburg's, which uses that structure so well. The second presentation doesn't tell you much about how to multiply elements, while with the first it is easy to see that you can push all the $c$s to the end and find a standard form for any given element. –  Mark Bennet Aug 17 '13 at 18:56
    
This is very useful answer and comment. I am lucky you didn't delete it. thank you –  Ronald Aug 17 '13 at 18:58

We can construct a directed graph called the Cayley diagram of the group $G := \langle a,b,c | a^2=b^2=c^3=1, ab=ba, ca=bc, cb=abc \rangle$ with respect to the generators $a,b,c$. The vertices of this digraph will be the group elements. The set of arcs will be of of the form $\{(g,gs): g \in G, s \in \{a,b,c\} \}$. After this digraph is constructed completely, each vertex $g \in G$ will have 3 outgoing arcs, labeled with colors $a,b,c$, and 3 incoming arcs with colors $a,b,c$, and the number of vertices will be the order of the group.

To construct this digraph, start with the identity vertex 1. Draw an arc from 1 to $a$, and label the arc as $a$. This is an edge-colored arc, denoted $(1,a,a)$, where the third coordinate refers to the color of the arc $(1,a)$. Add a new arc from vertex $a$ to vertex $a^2$ with color $a$, and notice that the relation $a^2=1$ forces us to identify the vertex $a^2$ and $1$. So we really have a directed cycle of length 2 at this point. Another directed cycle of length 2 arises due to arcs $(1,b,b)$ and $(b,1,b)$. Next we draw a directed cycle of length 3 on vertices $1,c,c^2$.

Next start at vertex $a$ and draw its 3 outgoing arcs, one of which will $(a,b,ab)$. Draw three outgoing arcs from $b$, one of which will be $(b,a,ba)$. Since $ab=ba$, we identify (i.e. merge) these two vertices $ab$ and $ba$. In addition, we start at the existing vertices (say the vertex 1), and enforce each relation at each vertex. Thus, starting at 1, the directed path $ab$ of length 2 and the directed path $ba$ of length 2 must end up at the same vertex. Similarly, $ca$ and $bc$ end up at the same vertex, as do $cb$ and $abc$.

Continue this process until every vertex has three outgoing and three incoming arcs, and until each relation is enforced at each vertex. We identify vertices whenever a relation forces us to do so. The number of vertices will then end up being the order of the group. I constructed this digraph in your example and verified it has order 12.

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Does this process also really prove that the order is exactly $12$ and not just $\leq 12$? –  Martin Brandenburg Aug 18 '13 at 8:04
    
@MartinBrandenburg It depends on what AG. means by "verified" in the last line. This gives us 12 group elements, and we need to verify that each are non-equal. –  user1729 Aug 18 '13 at 10:06
    
I believe this process does give the exact value of $|G|$ for any presentation. Let $F$ be the free group generated by the symbols $a,b,c, \ldots$, and let $N$ be the smallest normal subgroup of $F$ generated by the relators $r_1,r_2,\ldots$. (In the above example, $r_1=a^2, \ldots, r_6 = cbc^{-1}b^{-1}a^{-1}$). Then $G$ is defined to be the quotient group $F/N$. It can be shown that two words $w_1$ and $w_2$ are equivalent in $G$ iff $w_2$ can be obtained from $w_1$ by a sequence of insertions and deletions of the form $x x^{-1}, x^{-1} x, r_i, r_i^{-1}$, which the procedure accounts for. –  AG. Aug 18 '13 at 12:04
    
In other words, it suffices to count the number of vertices of the final digraph to obtain the order of the group. Distinct vertices of the final digraph would automatically correspond to distinct group elements, so no further verification is necessary. –  AG. Aug 18 '13 at 13:13

The presentation can be rewritten as $\langle d,c\mid d^2 =c^3 = (cd)^3=1 \rangle$. This is the standard presentation for the symmetry group (rotations) of the regular tetrahedron where $d$ represents a rotation about an edge, $c$ represents a rotation about a face and $cd$ represents a rotation about a vertex.

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