Skip to main content
typos
Source Link
Geoff Robinson
  • 44.4k
  • 5
  • 123
  • 169

$G$ is non-solvable, so must have $A_{5}$ as a composition factor (as no other non-Abelian simple group has less than $168$). Hence $F(G)$ can have order at most $4$.

If $G$ has no component, then $F(G) = F^{\ast}(G)$ has order $4$, and is centralized by all elements of ordderorder $5$, contrary to $C_{G}(F^{\ast}(G)) \leq F^{\ast}(G).$

Hence $G$ has a component (quasi-simple subnormal subgroup). This is isomorphic to either $A_{5}$ or ${\rm SL}(2,5).$ If it is (isomorphic to) ${\rm SL}(2,5),$ it is then normal (it has index two), and ${\rm SL}(2,5)$ has an irreducible character of degree $6$. By Clifford's theorem, $G$ has an irreducible character character of degree at least $6,$ which is precluded here.

Hence $G$ has a normal subgroup $L$ isomorphic to $A_{5}.$ Each irreducible character of degree $3$ of $A_{5}$ must extend to an irreducible character of degree $3$ of $G$ (since $G$ has no irreducible character of degree $6$ or more). Hence no element of order $5$ in $L$ can be conjugate to all its non-identity powers. Let $S$ be a Sylow $5$-subgroup of $L$. Then by a Frattini argument, $G = LN_{G}(S) = LC_{G}(S)$ (since $[N_{G}(S):C_{G}(S)] = [N_{L}(S):C_{L}(S)] = 2).$ Now $G$ induces inner automorphisms on $F^{\ast}(G) = L \times O_{2}(G)$ so $G = F^{\ast}(G)$ has one of the listed structures.

$G$ is non-solvable, so must have $A_{5}$ as a composition factor (as no other non-Abelian simple group has less than $168$). Hence $F(G)$ can have order at most $4$.

If $G$ has no component, then $F(G) = F^{\ast}(G)$ has order $4$, and is centralized by all elements of ordder $5$, contrary to $C_{G}(F^{\ast}(G)) \leq F^{\ast}(G).$

Hence $G$ has a component (quasi-simple subnormal subgroup). This is isomorphic to either $A_{5}$ or ${\rm SL}(2,5).$ If it is (isomorphic to) ${\rm SL}(2,5),$ it is then normal (it has index two), and ${\rm SL}(2,5)$ has an irreducible character of degree $6$. By Clifford's theorem, $G$ has an irreducible character character of degree at least $6,$ which is precluded here.

Hence $G$ has a normal subgroup $L$ isomorphic to $A_{5}.$ Each irreducible character of degree $3$ of $A_{5}$ must extend to an irreducible character of degree $3$ of $G$ (since $G$ has no irreducible character of degree $6$ or more). Hence no element of order $5$ in $L$ can be conjugate to all its non-identity powers. Let $S$ be a Sylow $5$-subgroup of $L$. Then by a Frattini argument, $G = LN_{G}(S) = LC_{G}(S)$ (since $[N_{G}(S):C_{G}(S)] = [N_{L}(S):C_{L}(S)] = 2).$ Now $G$ induces inner automorphisms on $F^{\ast}(G) = L \times O_{2}(G)$ so $G = F^{\ast}(G)$ has one of the listed structures.

$G$ is non-solvable, so must have $A_{5}$ as a composition factor (as no other non-Abelian simple group has less than $168$). Hence $F(G)$ can have order at most $4$.

If $G$ has no component, then $F(G) = F^{\ast}(G)$ has order $4$, and is centralized by all elements of order $5$, contrary to $C_{G}(F^{\ast}(G)) \leq F^{\ast}(G).$

Hence $G$ has a component (quasi-simple subnormal subgroup). This is isomorphic to either $A_{5}$ or ${\rm SL}(2,5).$ If it is (isomorphic to) ${\rm SL}(2,5),$ it is then normal (it has index two), and ${\rm SL}(2,5)$ has an irreducible character of degree $6$. By Clifford's theorem, $G$ has an irreducible character of degree at least $6,$ which is precluded here.

Hence $G$ has a normal subgroup $L$ isomorphic to $A_{5}.$ Each irreducible character of degree $3$ of $A_{5}$ must extend to an irreducible character of degree $3$ of $G$ (since $G$ has no irreducible character of degree $6$ or more). Hence no element of order $5$ in $L$ can be conjugate to all its non-identity powers. Let $S$ be a Sylow $5$-subgroup of $L$. Then by a Frattini argument, $G = LN_{G}(S) = LC_{G}(S)$ (since $[N_{G}(S):C_{G}(S)] = [N_{L}(S):C_{L}(S)] = 2).$ Now $G$ induces inner automorphisms on $F^{\ast}(G) = L \times O_{2}(G)$ so $G = F^{\ast}(G)$ has one of the listed structures.

typo
Source Link
Geoff Robinson
  • 44.4k
  • 5
  • 123
  • 169

$G$ is non-solvable, so must have $A_{5}$ as a composition factor (as no other non-Abelian simple group has less than $168$). Hence $F(G)$ can have order at most $4$.

If $G$ has no component, then $F(G) = F^{\ast}(G)$ has order $4$, and is centralized by all elements of ordder $5$, contrary to $C_{G}(F^{\ast}(G)) \leq F^{\ast}(G).$

Hence $G$ has a component (quasi-simple subnormal subgroup). This is isomorphic to either $A_{5}$ or ${\rm SL}(2,5).$ If it is (isomorphic to) ${\rm SL}(2,5),$ it is then normal (it has index two), and ${\rm SL}(2,5)$ has an irreducible character of degree $6$. By Clifford's theorem, $G$ has an irreducible character character of degree at least $6,$ which is precluded here.

Hence $G$ has a normal subgroup $L$ isomorphic to $A_{5}.$ Each irreducible character of degree $3$ of $A_{5}$ must extend to an irreducible character of degree $3$ of $G$ (since $G$ has no irreducible character of degree $6$ or more). Hence no element of order $5$ in $L$ can be conjugate to all its non-identity powers. Let $S$ be a Sylow $5$-subgroup of $L$. Then by a Frattini argument, $G = LN_{G}(S) = LC_{G}(S)$ (since $[N_{G}(S):C_{G}(S)] = [N_{L}(S):C_{L}(S)] = 2).$ Now $G$ induces inner automorphisms on $F^{\ast}(G) = L \times O_{2}(G)$ so $G = F^{\ast}(G)$ has one of the listed structures.

$G$ is non-solvable, so must have $A_{5}$ as a composition factor (as no other non-Abelian simple group has less than $168$). Hence $F(G)$ can have order at most $4$.

If $G$ has no component, then $F(G) = F^{\ast}(G)$ has order $4$, and is centralized by all elements of ordder $5$, contrary to $C_{G}(F^{\ast}(G)) \leq F^{\ast}(G).$

Hence $G$ has a component (quasi-simple subnormal subgroup). This is isomorphic to either $A_{5}$ or ${\rm SL}(2,5).$ If it is (isomorphic to) ${\rm SL}(2,5),$ it is then normal (it has index two), and ${\rm SL}(2,5)$ has an irreducible character of degree $6$. By Clifford's theorem, $G$ has an irreducible character character of degree at least $6,$ which is precluded here.

Hence $G$ a normal subgroup $L$ isomorphic to $A_{5}.$ Each irreducible character of degree $3$ of $A_{5}$ must extend to an irreducible character of degree $3$ of $G$ (since $G$ has no irreducible character of degree $6$ or more). Hence no element of order $5$ in $L$ can be conjugate to all its non-identity powers. Let $S$ be a Sylow $5$-subgroup of $L$. Then by a Frattini argument, $G = LN_{G}(S) = LC_{G}(S)$ (since $[N_{G}(S):C_{G}(S)] = [N_{L}(S):C_{L}(S)] = 2).$ Now $G$ induces inner automorphisms on $F^{\ast}(G) = L \times O_{2}(G)$ so $G = F^{\ast}(G)$ has one of the listed structures.

$G$ is non-solvable, so must have $A_{5}$ as a composition factor (as no other non-Abelian simple group has less than $168$). Hence $F(G)$ can have order at most $4$.

If $G$ has no component, then $F(G) = F^{\ast}(G)$ has order $4$, and is centralized by all elements of ordder $5$, contrary to $C_{G}(F^{\ast}(G)) \leq F^{\ast}(G).$

Hence $G$ has a component (quasi-simple subnormal subgroup). This is isomorphic to either $A_{5}$ or ${\rm SL}(2,5).$ If it is (isomorphic to) ${\rm SL}(2,5),$ it is then normal (it has index two), and ${\rm SL}(2,5)$ has an irreducible character of degree $6$. By Clifford's theorem, $G$ has an irreducible character character of degree at least $6,$ which is precluded here.

Hence $G$ has a normal subgroup $L$ isomorphic to $A_{5}.$ Each irreducible character of degree $3$ of $A_{5}$ must extend to an irreducible character of degree $3$ of $G$ (since $G$ has no irreducible character of degree $6$ or more). Hence no element of order $5$ in $L$ can be conjugate to all its non-identity powers. Let $S$ be a Sylow $5$-subgroup of $L$. Then by a Frattini argument, $G = LN_{G}(S) = LC_{G}(S)$ (since $[N_{G}(S):C_{G}(S)] = [N_{L}(S):C_{L}(S)] = 2).$ Now $G$ induces inner automorphisms on $F^{\ast}(G) = L \times O_{2}(G)$ so $G = F^{\ast}(G)$ has one of the listed structures.

Source Link
Geoff Robinson
  • 44.4k
  • 5
  • 123
  • 169

$G$ is non-solvable, so must have $A_{5}$ as a composition factor (as no other non-Abelian simple group has less than $168$). Hence $F(G)$ can have order at most $4$.

If $G$ has no component, then $F(G) = F^{\ast}(G)$ has order $4$, and is centralized by all elements of ordder $5$, contrary to $C_{G}(F^{\ast}(G)) \leq F^{\ast}(G).$

Hence $G$ has a component (quasi-simple subnormal subgroup). This is isomorphic to either $A_{5}$ or ${\rm SL}(2,5).$ If it is (isomorphic to) ${\rm SL}(2,5),$ it is then normal (it has index two), and ${\rm SL}(2,5)$ has an irreducible character of degree $6$. By Clifford's theorem, $G$ has an irreducible character character of degree at least $6,$ which is precluded here.

Hence $G$ a normal subgroup $L$ isomorphic to $A_{5}.$ Each irreducible character of degree $3$ of $A_{5}$ must extend to an irreducible character of degree $3$ of $G$ (since $G$ has no irreducible character of degree $6$ or more). Hence no element of order $5$ in $L$ can be conjugate to all its non-identity powers. Let $S$ be a Sylow $5$-subgroup of $L$. Then by a Frattini argument, $G = LN_{G}(S) = LC_{G}(S)$ (since $[N_{G}(S):C_{G}(S)] = [N_{L}(S):C_{L}(S)] = 2).$ Now $G$ induces inner automorphisms on $F^{\ast}(G) = L \times O_{2}(G)$ so $G = F^{\ast}(G)$ has one of the listed structures.