5. let H be a subgroup of a group G such that g ?¹ hg  Show every left
coset gH is the same as the right coset Hg

6. prove that if G is an abelian group, written multiplicatively, with
identity element e, then all elements, x, of G satisfying the equation
x²=e form a sub group H of G

 8. prove the generalisation of the first part of this question:
consider the set H of all solutions, x, of the equation x ? =e for
fixed integer n ?1 in an abelian group, G  with identity , e.

9. if ? is a binary operation on a set, S an element, x  idempotent
for ?  if x ? x= x. prove that a group has exactly one idempotent
element.

10. define the term ?normal subgroup?. Give an example of a group, G,
and a normal subgroup, H, of G

11. prove that every group, G, with identity, e, such that x ? x=e for
all x G is abelian .

12. draw the cayley tables for the Z  and V. for each group, list the
pairs of inverses

13. determine whether the following are hohmorphisms. Let: 
i. ?  : Z  ? R under addition be given by  ? (n) =n. 
ii let G be any group and let : G? G be given by ? (g)=g?¹ 
for g element ofG

Ah, I miss college math.  Perhaps I'll go back and finish those last 3
classes for the math major... Group Theory II is one of those.

5.  Hm... I'm not quite sure you've written out the whole question
here.  "let H be a subgroup of a group G such that g ?¹ hg"... such
that g ?¹ hg what?  I think I can help you with this one if you
clarify what you're asking.

6. Given: x²=e, G is abelian

and we want to show that all of these x's form a subgroup.  Well, what
is a subgroup?  It's a group within a group.  We already know it's
within the group... these x's are all elements of G.  So is H a group?

There are four things that make a group a group.  Let's look at them
piece by piece.
Is H associative?  It inherits this property from being in G, so yes.
Does it have an identity?  The identity of a subgroup would be the
same as the identity of the group, so e, in this case.  Is e in H? 
well, e²=ee=e, which is a property of the identity, so yes.  Good! 
We're halfway there.
How about inverses?  In order for an element in H to have an inverse,
there must first be an element in H.  Let's pick one and call it a.  a
is in H, so aa=e.  We also know that aa?¹=e, because this is a
property of inverses.  Let's play a bit...
e=ee=(aa?¹)(aa?¹)=aa?¹aa?¹=aaa?¹a?¹=(aa)(a?¹a?¹)=e(a?¹a?¹)=a?¹a?¹. 
We've just shown that (a?¹)²=e, and therefore a?¹ must be in H if a
is.
Is H closed?  We know a and b are in H, or that a²=e and b²=e.  What
about ab?  (ab)²=abab=aabb=(aa)(bb)=ee=e.  So if a and b are in H, ab
must be a well.
Hey!  That's it!  H is a subgroup!

8.  Again, this doesn't seem to give enough information.  Maybe you
have missed writing down some of the question?  As it stands, there is
nothing to prove in the statement.

9.  Rather than using your binary operator symbol, I hope you don't
mind if I just use juxtaposition.  If you really want the symbol
there, you can copy and paste the equations before looking at them,
and pasting your symbol between each element I've written.

If a group has exactly one idempotent element, we need to figure out
what would be wrong with a group trying to contain 2 or more
idempotents.  Let's again look at what makes a group:
Closure:  If we have two elements a and b such that aa=a and bb=b, it
seems that we're still getting things in the group.  I don't think
closure is our problem... we're not introducing a c that may not be in
the group.
Identity:  ee=e certainly, so I'm guessing this is our one idempotent
element no matter what.  But if aa=a also, I don't see how that
effects the existance of e in G, so this is probably not our problem
either.
Inverses:  Let's say a is our second idempotent (remember that e is
always our first).  e=aa?¹=(aa)a?¹=a(aa?¹)=ae=a.  Hm... we seem to be
down to just one idempotent again, as anything that we think is not
the identity turns out to be as well.  We'll see why this turns out to
be necessary in a moment, with...
Associativity:  If ab=c but aa=a, we should get that aab=c.  This is
fine if we associate (aa)b=ab=c, but what about a(ab)?  Then we get
that a(ab)=ac=c.  The only time an element leaves another element
unchanged is if that first element is the identity.  Otherwise, weird
things are going on.
So clearly, there can only be one idempotent, the identity, as
anything else just turns out to be the identity thinly disguised.

10.  A subgroup H is normal if for every g in G and h in H, g?¹hg
exists in H.  This should look very similar to problem 5.  This is
because normal subgroups are less rigorously defined as subgroups for
which the left cosets are equivalent to the right cosets.  I'll let
you find your own examples, as I don't know with what types of group
you're supposed to be familliar.

11. A group is abelian if ab=ba for all a and b in G.  We know that
aa=bb=e.  Note that since ab is an element of G, (ab)(ab)=e also. 
aabb=(aa)(bb)=ee=e=(ab)(ab)=abab.  Now let's manipulate this equation
aabb=abab.  a?¹aabb=a?¹abab, (a?¹a)abb=(a?¹a)bab, eabb=ebab, abb=bab,
abbb?¹=babb?¹, ab(bb?¹)=ba(bb?¹), abe=bae, ab=ba.  We are done.

12. Sorry... I haven't done Cayley tables.  I'll look it up and see if
I can explain it to you later though, if you'd like.

13. Well, to be a homomorphism, ?(ab)=?(a)?(b) (or in the case of
addition, it may be easier for you to see that the equation is
?(a+b)=?(a)+?(b)).
i. ?(n)=n?  Well, then ?(a+b)=(a+b), and since ?(a)=a and ?(b)=b,
?(a)+?(b)=a+b, so yes, this is a homomorphism.
ii.?(g)=g?¹?  Well, ?(a)?(b)=a?¹b?¹, but ?(ab)=(ab)?¹=b?¹a?¹, so this one's not.

I hope that was somewhat helpful.