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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(, p) (the question of what subgroups of GL(, p), the general linear group of degree 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 )