Re: I need help proving a theorem!!!

In article <1190923094.777497.159560@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<jastein4930@xxxxxxxxxxxxx> wrote:
Thm: Let G be a group and let H be a subgroup of G. Define N(H)=(x
exists in G such that xHx^-1=H). Prove that N(H), called the
normalizer of H, is a subgroup of G.

The only ways I have learned to show that a group is a subgroup under
another group is using the 1-step, 2-step, or finite subgroup tests.

What the heck are the "1-step", the "2-step", and the "finite
subgroup" tests?

A subset H of a gropu G is a subgroup if and only if it is a group
under the restriction of the operation from G. This reduces to
checking that it satisfies the following three conditions:

(i) the identity element of G is an element of H;
(ii) H is closed under multiplication (if x and y are in H, then xy
is in H);
(iii) H is closed under inverses (if x is in H, then x^{-1} is in H).

These three conditions are equivalent to:

(1) H is nonempty; and
(2) If x and y are in H, then xy^{-1} is in H.


Perhaps these are what you call "2-step" and "1-step" tests... I have
no idea what "finite subgroup" test might be, though.

Am I supposed to be using one of these tests to prove this?

You are supposed to be checking that the set in question is a
subgroup. I.e., that it satisfies the definition. You can either do
it directly, or you can do it applying theorems you know that say "If
suchandsuch is true, then H is a subgroup", in which case you would
have to verify and establish that suchandsuch is indeed true.

So for example, you would want to check that the set satisfies (i),
(ii) and (iii) above. Or you would want to check that it satisfies (1)
and (2). Or some other conditions you know guarantee a set is a
subgroup.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

.

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