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Direct Products, Semidirect Products, and How to Visualize Them

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Direct Products, Semidirect Products, and How to Visualize Them Visualizing Groups Notes: Day 2 Sara Fish In today's class, we will cover: direct products, semidirect products, and how to visualize them. We will start by doing a lot of technical stuff. Then pictures come at the end. To motivate this discussion, consider D8 and Z2 × Z4. They have very similar Cayley graphs: (1; 3) (1; 2) r3s r2s (0; 3) (0; 2) r3 r2 e r (0; 0) (0; 1) s rs (1; 0) (1; 1) D8 Z2 × Z4 It would be nice to be able to say that both are \built from" Z2 and Z4. It turns out that we can write D8 = Z4 o Z2, where o is the symbol for semidirect product. We will slowly build up what this means. Direct products Definition. Let H and K be groups. Then we define H × K := f(h; k)j h 2 H; k 2 Kg to be the external direct product of H and K, with multiplication rule (h; k) · (h1; k1) = (hh1; kk1): Proposition. 1. H × K is a group. 2. jH × Kj = jHjjKj. 3. H; K E H × K, where H and K are identified with their isomorphic copies H × f1K g and f1H g × K in H × K. Proof. The external direct product isn't really that exiting. We're just taking two groups and putting them side by side. The internal direct product is slightly more interesting: Definition. Let G be a group. Let H; K E G such that H \ K = feg. Then HK = fhkj h 2 H; k 2 Kg is called the internal direct product. Proposition. 1. The groups H and K commute. That is, for all h 2 H and k 2 K, we have hk = kh. 2. HK is a group with multiplication hk · h1k1 = (hh1)(kk1). Proof. 1 Visualizing Groups Notes: Day 2 Sara Fish Remark. Note that the internal direct product and the external direct product are basically the same thing. It's just different notation. (h; k) · (h1; k1) = (hh1; kk1) (external direct product) hk · h1k1 = (hh1)(kk1) (internal direct product) Formally, we say that H × K and HK are isomorphic. Note that this isomorphism only works because of the precise way we defined HK. Semidirect products Suppose that we have two subgroups N; H ≤ G such that N \ H = feg, but this time, only one of them is normal: N E G. Constructing a internal direct product doesn't work anymore since H is not a normal subgroup. We salvage this construction as follows. Proposition. NH = fnhj n 2 N; h 2 Hg is a subgroup. Proof. Definition. Let N; H ≤ G, N E G, and N \ H = feg. Then NH is defined to be the internal semidirect product, which is a group with multiplication −1 nh · n1h1 = n(hn1h )hh1: Next we define the external semidirect product. This is going to be most useful to use, because it will allow us to construct bigger groups from G = N o H from unrelated groups N and H without knowing G in advance. Recall that for direct products, the internal and external direct products were the same thing. We want this to be true for semidirect products as well. Here's the problem. If we simply write down the multiplication rule using a few more parentheses and commas we get −1 (n; h) · (n1; h1) = (n(hn1h ); hh1): But this isn't a valid construction of an external product. For the external semidirect product, we will be multiplying two groups N and H which are completely unrelated. So we cannot compute −1 hn1h , since that involves multiplying elements from different groups. Here's the trick. We don't need arbitrary multiplication of elements in N and H. All we need is conjugating some n 2 N by h 2 H, that is, hnh−1. Since N is a normal subgroup, conjugating by some h 2 H, i.e. the map n 7! hnh−1 for all n 2 N, is an automorphism of N. Every h 2 H will induce an automorphism of N, given by conjugation by that h. So if we additionally specify a φh 2 Aut(N) for every h 2 H, then we can define the external semidirect product. Definition. Let N and H be groups. Let Φ : H ! Aut(N) be a homomorphism. We write Φ(h) = φh. Then we define the exteral semidirect product N oΦ H to be the set N × H with multiplication rule (n; h) · (n1; h1) = (nφh(n1); hh1): Proposition. 2 Visualizing Groups Notes: Day 2 Sara Fish 1. N oΦ H is a group. 2. N E N oΦ H and H ≤ N oΦ H, where N and H are identified with their isomorphic copies in N oΦ H. 3. The isomorphic copies of N and H in N oΦ H satisfy N \ H = feg. −1 4. Each φh is conjugation by h. That is, (1; h)(n; 1)(1; h ) = φh(n). Proof. Remark. Because we were careful, the internal semidirect product and the external semidirect product are basically the same thing. Again, it's just different notation: (n; h) · (n1; h1) = (nφh(n1); hh1) (external semidirect product, where φh is conjugation by h) −1 nh · n1h1 = n(hn1h )hh1 (internal semidirect product) Formally, we say that NH and N oΦ H are isomorphic. Remark. Note that the direct product is a special case of the semidirect product. This is because if Φ = id, then N oΦ H = N × H. Visualizations Remark. We will stop writing vertex labels. Last class, we proved that Cayley graphs \look the same" at every point. Since there is a Cayley graph isomorphism from g to g0 for any two g; g0 2 G, Cayley graph labelings are largely arbitrary. We can pick an arbitrary vertex to be the identity and “fill it in" from there. Example. r2 r3 r2s r3s arbitrarily set e fill in the rest −! −! rs s e r e Direct products Example. Consider the Cayley graph of Z4 ×Z2. Note that it can be seen as 2 copies of Z4 stitched together by copies of Z2. Or it can be seen as 4 copies of Z2 stitched together by copies of Z4. −! −! Z2 Z4 Z2 × Z4 Z4 × Z2 Put a copy of Z4 at each vertex of Z2. Put a copy of Z2 at each vertex of Z4. 3 Visualizing Groups Notes: Day 2 Sara Fish More generally, we can draw the Cayley graph of H × K as follows: 1. Draw the Cayley graph of K. 2. Replace each vertex k 2 K with the coset Hk. 3. Connect corresponding vertices (i.e. vertices with the same value of h). Example. Z2 × Z2 −! −! Example. Z3 × Z4 −! −! Example. Z2 × Z2 × Z2 −! −! Remark. You might say to yourself: \These examples aren't interesting because they're just direct products of cyclic groups, which are easy to understand." You're right, but it's not because I am bad at picking examples. In fact, by the fundamental theorem of finitely generated abelian groups, every finite abelian group can be expressed as a direct product Zk1 × · · · × Zkn . So once we understand direct products of cyclic groups, we understand all finite abelian groups! Semidirect products Now we turn to the general case of drawing the Cayley graph of N oΦ H. Definition. A rewiring of Γ(N; S) by h 2 H is a graph in which: 1. Replace each vertex n 2 N with φh(n). 2. Draw in edges such that the resulting graph is a Cayley graph. Example. The rewiring of Γ(Z4; frg) by f 2 Z2 can be computed as follows: 4 Visualizing Groups Notes: Day 2 Sara Fish 3 2 2 r3 r2 fr f = r fr f = r −! e r fef = e frf = r3 Before rewiring After rewiring Remark. In general, here is how to draw the Cayley graph of N oΦ H with generating set S [ T , where S generates N and T generates H. 1. Draw the Cayley graph of H. 2. At each vertex h 2 H, replace h with the rewiring of Γ(N; S) by h. 3. Connect corresponding vertices from each rewiring. Example. Let's draw the Cayley graph of Z4 oΦ Z2. Suppose Z4 = hri and Z2 = hfi. First we need to determine what Φ could be. In this case, Φ : Z2 ! Aut(Z4). What even is Aut(Z4)? There 3 are two automorphisms of Z4, namely r 7! r and r 7! r . (Note that since Z4 is cyclic, it suffices to specify the image of r.) Consider Φ again. Since Φ is a homomorphism, it has to map the identity to the identity, so it maps e to r 7! r. As for f, Φ can map f to either r 7! r or r 7! r3. If Φ maps f to r 7! r, then Φ = id, so Z4 oΦ Z2 = Z4 × Z2. This is just a direct product. So the only \actual" semidirect product is the case where Φ maps f to r 7! r3. Now we can draw the Cayley graph of Z4 oΦ Z2. First, draw the Cayley graph of Z2: e f Next, replace each h 2 Z2 with the rewiring of Γ(Z4; r) by h: Finally, connect corresponding vertices and rearrange: −! Voila! The Cayley graph of D8. 5 Visualizing Groups Notes: Day 2 Sara Fish Example. Let's draw the Cayley graph of A4 = V4 o Z3. Consider V4 with generating set f(12)(34); (13)(24)g and Z3 with generating set f(123)g.
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