# A few Abstract Algebra questions

• Feb 3rd 2009, 09:17 PM
Janu42
A few Abstract Algebra questions
I have a few problems I can't figure out, I'm sure they're not too complex, I just can't grasp how I'm supposed to tackle these.... (* means binary operation in these)

1. Let (G, *) be a group and let g be some fixed element of G. Show that G = {g*x such that x is in G}

2. Let G be a nonempty set and let * be an associative binary operation on G. Assume that both the left and right cancellation laws hold in (G, *). Assume moreover that G is finite. Show that (G, *) is a group. (Obviously, * being associative is a part of what makes it a group. However, I don't know how to show how an identity exists and an inverse. I don't get how the group being finite has anything to do with this.)

3. Let G be a group and let x in G be an element of order 18. Find the orders of x^2, x^3, x^4, x^5, x^12.

4. Show that if G is a finite group, then every element of G is of finite order.

5. Give an example of an infinite group G such that every element of G has finite order.
• Feb 4th 2009, 12:58 PM
Janu42
Anyone? For the first one, I talked to my professor, and said I need to prove each was a subset of the other. So I think I can get that one. But if someone can help me just one or a couple of the questions, that would be a big help? I'm pretty confused.
• Feb 4th 2009, 03:57 PM
ThePerfectHacker
Quote:

Originally Posted by Janu42
2. Let G be a nonempty set and let * be an associative binary operation on G. Assume that both the left and right cancellation laws hold in (G, *). Assume moreover that G is finite. Show that (G, *) is a group. (Obviously, * being associative is a part of what makes it a group. However, I don't know how to show how an identity exists and an inverse. I don't get how the group being finite has anything to do with this.)

Let $\displaystyle G = \{x_1,...,x_n\}$ and consider the function $\displaystyle \phi_g: G\to G$ defined by $\displaystyle \phi_g(x) = gx$. This function is one-to-one, because if $\displaystyle \phi_g(x) = \phi_g(y) \implies gx = gy \implies x=y$ by the left-cancellation law. But since $\displaystyle G$ is finite any injective map to itself must automatically be a surjective map (pigeonhole principle). Therefore, for any $\displaystyle b\in G$ there exists $\displaystyle x\in G$ so that $\displaystyle \phi_g(x) = b$ i.e. $\displaystyle gx = b$. We have established that the equation $\displaystyle gx = b$ is always sovable (for $\displaystyle x$) for any $\displaystyle g,b\in G$. Similarly, $\displaystyle xg=b$ is always solvable. Pick any $\displaystyle g\in G$. We have established that there exists $\displaystyle e\in G$ so that $\displaystyle ge=g$. We will argue now that $\displaystyle e$ is a right identity element. Let $\displaystyle b\in G$ so there is $\displaystyle x$ such that $\displaystyle xg=b$. Thus, $\displaystyle be = (xg)e = x(ge) = xg = b$. Great, we have found a left identity element. Now find a right identity element, show they are the same and fill in all the steps.
• Feb 4th 2009, 04:07 PM
ThePerfectHacker
Quote:

Originally Posted by Janu42
3. Let G be a group and let x in G be an element of order 18. Find the orders of x^2, x^3, x^4, x^5, x^12.

If $\displaystyle x$ has order $\displaystyle 18$ then $\displaystyle x^k$ has order $\displaystyle \tfrac{18}{(k,18)}$.

Quote:

4. Show that if G is a finite group, then every element of G is of finite order.
Let $\displaystyle G = \{x_1,x_2,...,x_n\}$.
If $\displaystyle x = e$ then proof complete.
If $\displaystyle x\not = e$ and $\displaystyle x^2=e$ then proof complete.
If $\displaystyle x,x^2\not = e$ and $\displaystyle x^3 = e$ then proof complete.
And keep on going.

Say that $\displaystyle x,x^2,x^3,...,x^k$ are all non-identity elements. Then all of these must be distinct. Because otherwise $\displaystyle x^i = x^j \implies x^{i-j} = e$ (assuming $\displaystyle i>j$) and this is a contradiction. The point here is if $\displaystyle x$ is a non-identity element all exponents of $\displaystyle x$ that are not idenitity elements (i.e. $\displaystyle x,x^2,...,x^k$) must be distinct. However, $\displaystyle G$ is finite and so we eventually reach a point when the number of exponents will exceede the number of elements in the group. In this case eventually the exponent produces the identity element.

Quote:

5. Give an example of an infinite group G such that every element of G has finite order.
Let $\displaystyle G= \{(x_1,x_2,x_3,... ) : x_i = 0 \text{ or } 1\}$.
Define $\displaystyle (x_1,x_2,...) + (y_1,y_2,...) = (z_1,z_2,...)$ where $\displaystyle z_i = x_1 + x_2 (\bmod 2)$.
Check that $\displaystyle G$ is a group.
Now prove that all non-trivial elements have order $\displaystyle 2$. (Surprised)
• Feb 4th 2009, 04:32 PM
Janu42
Thank you so much. That was a big help and I really appreciate it.