Determine whether each of the following statement is true or false. If it is tru

Question

Determine whether each of the following statement is true or false. If it is true prove it. If it is false explain why or give an example that disproves the statement. The mapping from Zn to Z by a rightarrow a is a group homoniorphism. If beta is an odd permutation, then beta-1 is odd. If G is a finite group and m divides the order of G. then there exists a subgroup H of G of order m. In a group of prime order, any subgroup is normal. If K is a normal subgroup of a group G and H is subgroup of G, then K H is a normal subgroup of H.

Explanation / Answer

2) true

the composition of two even permutations is even
the composition of two odd permutations is even
the composition of an odd and an even permutation is odd
From these it follows that

the inverse of every even permutation is even
the inverse of every odd permutation is odd

3)false in case of non abellion cases.

4)true

Let G have prime order p. Then by Lagrange's theorem, subgroups either have order 1 (identity subgroup; clearly normal) or p (=G), so it suffices to show that G is normal in G (that is, for all g in G, gG = Gg). Well, again by Lagrange's theorem, there is only one coset of G in G, so gG = G = Gg. (Alternatively, note that all prime order groups are cyclic, and normalcy follows easily).

5) True

That H ? K is a subgroup of G is easy to verify.

(This should have been an exercise you have done previously. Just in case...

Closure: Let x,y be in H ? K.
Then, x,y are in H and x,y is in K.
Since H and K are subgroups, xy is in H and xy is in K.
Thus, xy is in H ? K.

Inverses: Let x be in H ? K.
Then, x is in H and x is in K.
Since H and K are subgroups, x^(-1) is in H and x^(-1) is in K.
Thus, x^(-1) is in H ? K.)
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To show normality:
Let x in H ? K ==> x is in H and x is in K.

Then, for any g in G,
g^(-1) x g is in H because H is normal in G and x is in H.
and g^(-1) x g is in K because K is normal in G and x is in K.

Therefore, for any g in G,
g^(-1) x g is in H ? K.

Hence, H ? K is a normal subgroup of G, as required

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