please help me please solve this abstract algebra.. thank you?

Question

Show that if G is a finite group with identity e and with an even number of elements, then there is a ≠ e in g such that a* a = e.

Answer 1

I already answered this question for another poster,
but I'll repeat here for you.
Look at G - {e}.
We know this has an odd number of elements.
Take any element a and pair it with its inverse.
If the inverse of a equals a, we are done.
Else multiply a by it's inverse and delete the pair.
There are still an odd number of elements left.
Now consider the remaining set and do the same thing.
Now G is finite so a finite number of steps we
will have (in the worst case) only one element left.
But now we can only pair a with itself, so
a = a^-1 or a*a = e.
In number theory, this is the same strategy
we use to prove Wilson's theorem:
If p is prime then (p-1)! = -1(mod p).
Let's look at U_11, for example
The elements are
1 2 3 4 5 6 7 8 9 10
Let's consider the elements 2, ...10 and do the above
pairing.
Working mod 11 we delete the pairs
(2,6), (3,4), (5,9) and (7,8)
Only 10 is left so 10 is its own inverse.
So if you multiply all the elements together
and use this strategy, their product is 10 = -1(mod 11

Answer 2

one million and a couple of are distinctly a lot with the help of definition: H is, as you gave it, a classic subgroup. enable x be an ingredient of the coset gH. So x = gh for some h in H. Now, all of us comprehend that ok=ghg^{-one million} is likewise in H with the help of the normality of H. so which you would be able to write h=g^{-one million}kg, and so x = gg^{-one million}kg = kg the place ok is in H. So x is interior the suited coset Hg. we've subsequently shown that gH is a subset of Hg. the comparable argument may be utilized to teach that Hg is a subset of gH, so the two are equivalent. Now for the communicate, assume gH = Hg. enable h be an ingredient in H. Then gh is and ingredient of gH, and subsequently an ingredient of Hg. So we've that gh = kg for some ok in H. hence ghg^{-one million} = ok is in H. So H is ordinary. for 3: for section a you may desire to teach that ~ satisfies the residences of an equivalence relation: reflexive, transitive, symmetric. Reflexive: enable h and ok be the p.c. out ingredient in G, that are unavoidably interior the subgroups H and ok. Symmetric, if a = hbk, then b = h^{-one million}ak^{-one million}. because of the fact that H and ok are sub communities, h^{-one million} and ok^{-one million} are components of H and ok respectively, so a~b implies b~a. Transitive, if a = hbk and b = xcy the place h,x in H and ok,y in ok, then a = hxcyk. lower back, because of the fact that H and ok are subgroups, hx and yk are components of H and ok, so a~c. For b, nicely, i will assist you to artwork out the define your self.

Answer 3

then there is a ≠ e in g
Which g? You didn't define g.