Give an example of a group G with a normal subgroup H which is not 4 chat at-terisik subgroup. (2) Let 0 be a group. Recall that the Antrim subgroup 0(C7; is the intersection 4,1 all maximal subgroups...

SOLUTION
Solution 1:-
We know that every characteristic subgroup is normal subgroup, but the vice
versa is not the same that is there are normal subgroup which are not a
characteristic subgroup.
The above fact can be cleared from the following example:-
1) Given G a group and H being a normal subgroup.
Now let us suppose,
G is the H * H (direct product).
In this case, both of the subgroups

{1} * H & H * {1} are normal, but none of these is characteristic.
None of these above mentioned subgroups is invariant beneath the
automorphism (a, b) → (b, a) that swaps both factors
SOLUTION
Solution: 2:-
We have to prove that Φ (G) is precisely the set of all non generators of G. So
lets us now assume that g be a non-generator of G and presume that g is not in
the Φ (G).
It means that there will be minimum one maximal subgroup G which does say
N. So N is lesser than (g, N), this thing implies that G = since N is
maximal. So by the property of non generator: G = N, and it is contradiction.
because of its definition, maximal subgroups are proper.
Now lets us take converse case:-
Assume that g belong Φ (G) and assume that:
G = , but G is not = to .
So definitely must be enclosed in certain maximal subgroup of G, let us
suppose N. But g belongs to Φ (G) <= N, so G = = N, which is another
contradiction.
Solution 3 a) As H G, Hk is a subgroup of G. Also, since H G, H Hk.
So, this thing implies that Hk/H - solvable and H also solvable.
Now Let us suppose that G’ = Hk/H.
We see that Hk/H is approximately to K/H ∩ K. Now...