Best Answer:
Let |G|=35=5*7.

By Sylow's theorem, G has a Sylow 5-subgroup, P, and a Sylow 7-subgroup, Q.

Then |P|=5 and |Q|=7.

Furthermore, the number of Sylow 5-subgroups = n5 = 1 (mod 5), and n5 divides 7.

That means n5=1.

And the number of Sylow 7-subgroups = n7 = 1 (mod 7), and n7 divides 5.

Hence n7=1.

So there is exactly one subgroup of G of order 5, namely P,

and there is exactly one subgroup of G or order 7, namely Q.

Observe that P and Q intersect trivially, since elements of P have order 5 and elements of Q have other 7 (the order of an element divides the order of a group containing it).

Therefore there are 4+6=10 non-identity elements in P and Q together.

So P and Q together account for 11 elements of G, including the identity,

Since |G|=35>11, there exists some g that lies in G but neither in P nor Q.

The possible values for |g| are 1, 5, 7, and 35 by Lagrange's theorem, since the order of an element divides the order of the group.

Clearly |g| is not 1, since g cannot be the identity, because the identity is in P and Q.

And |g| cannot be 5, because otherwise g would generate a group of order 5, i.e., a Sylow 5-subgroup - a different Sylow 5-subgroup from P, because x is not in P by assumption. But that is impossible, because we showed that there is exactly 1 Sylow 5-subgroup of G.

Similarly |g| cannot be 7.

The only other possibility is that |g|=35.

But then g generates the entire group G.

So G is cyclic.

Therefore G is abelian, i.e., its center is the whole group G.

Source(s):

Anonymous
· 7 years ago

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