A characterization of L 3 (4) by its character degree graph and order

Proof

We know from Conway et al. (1985, p. 23), that ({mathrm {cd}}(L_3(4))={1, 20,35,45,63,64}). So the graph (Gamma (G)) is the graph with vertex set ({2,3,5,7}) and the vertices 5 and 7, and the vertices 2 and 7 have no edge. Therefore there is a character (chi in {mathrm {Irr}}(G)) with (5cdot 7mid chi (1)).

It is easy to prove that (O_5(G)=1) and (O_7(G)=1). In fact, if (O_7(G)ne 1), then (O_7(G)) is a normal abelian Sylow 7-subgroup of G of order 7 by hypotheses. Then by Lemma 1, (chi (1)mid |G:O_7(G)|) for all (chi (1)in {mathrm {cd}}(G)), a contradiction. Similarly we can prove that (O_5(G)=1).

Suppose first that G is soluble. Let (Mne 1) be a minimal normal subgroup of G. Then M is an elementary abelian p-group with p = 2 or 3. Note that (|G|_p = p) for p = 5, 7 and in (Gamma (G)), there is a character (chi ) of G such that (5cdot 7) divides (chi (1)). Then by Lemma 1, p = 2 or 3. So two cases are considered.

1. 1.

   Let M be a 3-group.

   Since there is a character (chi ) with (chi (1)=21), then (|M|=3). Let H / M be a Hall subgroup of G / M of order (2^6cdot 5cdot 7). Then (|G/M:H/M|=3). It follows that ((G/M)/(L/M)hookrightarrow S_3), where (S_3) is the symmetric group of degree 3 and (L/M={mathrm {Core}}_{G/M}(H/M):=bigcap _{gMin G/M}(H/M)^{gM}), the core of H / M in G / M. So we have (|L/M|mid 2^6cdot 5cdot 7) and (Q/Munlhd L/M), where (Q/Min {mathrm {Syl}}_p(L/M)) with p = 5 or 7. Hence since (L,{mathrm {ch}},G), (Qunlhd G ) and so (|Q|=3cdot p). Therefore (O_p(G)ne 1) with p = 5 or 7, a contradiction.

2. 2.

   Let M be a 2-group.

   If (|M|=2^6), then by Lemma 1, (chi (1)mid |G:M|), a contradiction. Hence (|M|=2^k) with (1le kle 5). Let H / M be a Hall subgroup of order (3^2cdot 5cdot 7). Then (|G/M:H/M|=|G:H|=2^kle 32).

   1. 2.1.

      If (1le kle 2), then (G/H_Ghookrightarrow S_{2^k}) and so (7mid |H_G|). Let (Q/Min {mathrm {Syl}}_7(H/M)). Also (|H_G/M|mid |H/M|=3^2cdot 5cdot 7). If (|H_G/M3^2cdot 5cdot 7), then (Q/M,{mathrm {ch}},H_G/Munlhd G/M) and so (Qunlhd G). It follows that (G_7) is normal in G, a contradiction. Hence (|H_G/M|=|H/M|=3^2cdot 5cdot 7). By hypotheses, we can choose a character (chi in {mathrm {Irr}}(G)) with (chi (1)=35). Let (theta in {mathrm {Irr}}(H)) with (e=[chi _H,theta ]ne 0). Then (35=ettheta (1)) with (t=|G:I_G(theta )|). Since the numbers e and t are divisors of (|G:H|=2^{6-k}), then (e=t=1) and so (chi _H=theta ) by Lemma 2. Since (theta (1)^2=5cdot 7cdot 5cdot 7|H|=2^kcdot 3^2cdot 5cdot 7) and (1le kle 2), then k = 2 and (|M|=4), (Msubseteq H). Let (eta in {mathrm {Irr}}(M)) such that (e’=[theta _M,eta ]ne 0). Therefore (35=e’t’) with (t’=|H:I_H(eta )|) Also we know that M has 4 linear characters and so (t’le 4). So ((e’,t’)=(35,1)). It follows that (35^2le [theta _M,theta _M]=e’^2t’le |H:M|=3^2cdot 5cdot 7), a contradiction.

   2. 2.2.
      If (3le kle 5), then (Mle H_Gle H). Therefore (pi (H_G)={2,3}), ({2,5}), ({2,7}), ({2,3,5}), ({2,3,7}), ({2,5,7}) or ({2,3,5,7}).
      1. 2.2.1.

         Let (pi (H_G)={2,3}).

         Since there is no character (chi ) with (6mid chi (1)) and M is abelian, then (|{mathrm {cd}}(H_G)|=2) and (3mid chi (1)) for some character (chi in {mathrm {Irr}}(H_G)). It folows from Isaacs (1994, Theorem 12.5), G either has an abelian normal subgroup of index 3 or 9 or is the product of a 3-group K and an abelian. If the former, k = 4 (otherwise, the Sylow 3-subgroup is normal in (H_G)) and (|K|=3). It means (3cdot 2^2mid |H/H_G|) and (2cdot 3in {mathrm {cd}}(G)), a contradiction. If the latter, then (H_G=(Z_3rtimes Z_3)times M) and so (3in {mathrm {cd}}(H_G)). It follows that there is also an character (chi ) such that (2cdot 3mid chi (1)), a contradiction.

      2. 2.2.2.

         Let (5in pi (H_G)).

         Let (Q/Min {mathrm {Syl}}_5(H_G/M)). Then (Qunlhd H_G,{mathrm {ch}},G) and so (G_5unlhd G), a contradiction.

      3. 2.2.3.

         Let (7in pi (H_G)).

         Let (Q/Min {mathrm {Syl}}_7(H_G/M)). Then (Qunlhd H_G,{mathrm {ch}},G) and so (G_7unlhd G), a contradiction.

      4. 2.2.4.

         Let (5,7in pi (H_G)).

         We can rule out this case as Case 2.2.2 or Case 2.2.3.

      5. 2.2.5.

         (H_G) is a 2-group.

         Then (M=H_G). If G is an elementary abelian, then by Webb (1983, p. 238), ({mathrm {Aut}}(G)) is an extension of an elementary abelian p-group of rank (frac{n^2(n-1)}{2}) by a subgroup of GL(n, p) (the question of what subgroups of GL(n, p), the general linear group of degree n over finite field of order p, can arise in this way is still far from a solution). We know that (frac{G}{C_G(M)}=frac{N_G(M)}{C_G(M)}lessapprox {mathrm {Aut}}(M)). If k = 3, then (5mid |C_G(M)|) and the Sylow 5-subgroup (G_5) of G is also a Sylow 5-subgroup of (C_G(M)). So (G_5unlhd G), a contradiction. if k = 4 or 5, then (7mid |C_G(M)|). Similarly we have (G_7unlhd G), a contradiction.

Therefore G is insoluble and so by Lemma 3, G has a normal series (1unlhd Hunlhd Kunlhd G), such that K/H is a direct product of isomorphic non-abelian simple groups and (|G/K|mid |{mathrm {Out}}(K/H)|).

We will prove that (5,7in pi (K/H)). Assume the contrary, then obviously by Kondrat’ev and Mazurov (2000, Lemma 6(d)) and Liu (2015, Lemma 2.13) apply to almost simple groups (K/Hle G/H le {mathrm {Aut}}(K/H)), where K/H has a disconnected prime graph. We have that (|{mathrm {Out}}(K/H)|) is divisible by neither 5 nor 7. If (5,7mid |H|), then there is a Hall ({5,7})-subgroup L of H, then L is cyclic and so L is abelian. By Lemma 1, (chi (1)mid |G:L|), a contradiction. If 5 divides the order |H| but (7nmid |H|), then (G_5) is cyclic and so get a contradiction by Lemma 1. Similarly, (7nmid |H|).

Therefore by Lemma 5 and considering group orders, K/H is isomorphic to one of the simple groups: (A_7), (A_8) or (L_3(4)).

If (K/Hcong A_7), then (A_7le G/Hle {mathrm {Aut}}(A_7)). If (G/Hcong A_7), then there is an edge between the vertices 2 and 3 in (Gamma (G)), a contradiction since ({mathrm {cd}}(A_7)={1, 6, 10, 14, 15, 21, 35}). Similarly, we can rule out when (G/Hcong S_7).

If (K/Hcong L_3(4)), then (L_3(4)le G/Hle {mathrm {Aut}}(L_3(4))). If (G/Hcong L_3(4)), then (H=1) and so (Gcong L_3(4)). For the other cases, we rule out by considering their orders.

If (K/Hcong A_8), then (A_8le G/Hle S_8). If (G/Hcong A_8), then (H=1) and so (Gcong A_8). But (Gamma (L_3(4))) has no edge between the vertices 2 and 7, a contradiction. If (G/Hcong S_8), then order consideration rules out.

So G is isomorphic to (L_3(4)).

This completes the proof. (square )