A Characterization of Sporadic Janko Group J 1

If n is an integer, then we denote by ( ) the set of all prime divisors of n. Let G be a group. The set of element orders of G is denoted by (G). Let k ∈ (G) and Sk be the number of elements of order k in G. Let nse (G) = {Sk| k ∈ (G)}. Let ( ) denote the set of prime p such that G contains an element of order p. A finite group G is called a simple Kn -group, if G is a simple group with | ( )| = n. Thompson posed a very interesting problem related to algebraic number fields as follows Shi (1989).


INTRODUCTION
If n is an integer, then we denote by ߨ(݊) the set of all prime divisors of n.Let G be a group.The set of element orders of G is denoted by ߱(G).Let k ∈ ߱(G) and S k be the number of elements of order k in G. Let nse (G) = {S k | k ∈ ߱(G)}.Let ‫)ܩ(ߨ‬ denote the set of prime p such that G contains an element of order p.A finite group G is called a simple Kn -group, if G is a simple group with ‫|)ܩ(ߨ|‬ = n.Thompson posed a very interesting problem related to algebraic number fields as follows Shi (1989).
Thomson' problem: Let T(G) ={(n, s | n ∈ ω(G) and S n ∈ nse (G)}, where S n is the number of elements with order n.Suppose that T(G) = T (H).If G is a finite solvable group, is it true that H is also necessarily solvable?
Comparing the sizes of elements of same order but disregarding the actual orders of elements in T (G) of the Thompson Problem, in other words, it remains only nse (G), whether can it characterize finite simple groups?Up to now, some groups especial for PSL (2,q), can be characterized by only the set nse (G) (Khalili et al., 2011;Shen et al., 2010), respectively).
The author has proved that the groups L 3 (4) and L 2 (16) are characterizable by nse (Liu, 2012b(Liu, , 2013) ) respectively).In this study, it is shown that the group J 1 , which the number of the set of the same order is 10, also can be characterized by nse (J 1 ), that is.

MAIN THEOREM
Let G be a group with nse (G) = nse (J 1 ).

Some lemmas:
Lemma 1: Frobenius (1895) Let G be a finite group and m be a positive integer dividing |G|.If , then Lemma 2: Miller (1904) Let G be a finite group and be odd.Suppose that P is a Sylow psubgroup of G and n = p 5 m with (p,m) = 1.If P is not cyclic and s>1, then the number of elements of order n is always a multiple of P s .
Lemma 3: Shen (2010) Let G be a group containing more than two elements.If the maximal numbers of elements of the same order in G is finite, then G is finite and Lemma 4: Hall (1959) [Theorem 9.3.1])Let G be a finite solvable group and |G| = mn, where , (m,n) = 1.Let and h m L be the number of Hall ߨ -subgroups of G. Then satisfies the following conditions for all :

•
(mod ) for some P j • The order of some chief factor of G is divided by

PROOF OF THEOREM
Let G be a group such that nse (G) = nse (J 1 ) and s n be the number of elements of order n.By Lemma 3 we have G is finite.We note that ( ) , where k is the number of cyclic subgroups of order n.Also we note that if n>2, then ( ) n , then by Lemma 1 and the above discussion, we have: (1) We rewrite the main theorem here to read easily.
The proof is the same as Case c.
The proof is the same as Case d.
We know that and .
We first show that and .
If 5.7∈ω (G), set P and Q are Sylow 7-subgroups of G, then P and Q are conjugate in G and so and are conjugate in G. Therefore we have that , where k is the number of cyclic subgroups of order 5 in .As , 100320 and then for some integer t, but the equation has no solution in N, a contradiction.Hence .It follows that the group P 5 acts fixed point freely on the set of order 7 and so (= 25080).Hence = 5.
Similarly since , we have that the groupP11 acts fixed point freely on the set of order 19 and so ( = 27720).Hence = 11.
Since , the group acts fixed point freely on the set of order 19 and so ( = 27720).