BG Math [Algebra sucks :(]

[SDSD]

I finished most of my review sheet for our exam but I just don't know where to start with this problem on subgroups. I'd appreciate if someone could point me in the right direction, and possible give me some tips on how to think about groups (since they seem to be my weak point).

14) Let G be a group and Z(G)= {z in G : z*x=x*z for all x in G}. Prove that Z(G) is a subgroup of G. Z(G) is called the center of G. If G is abelian, what is Z(G)?

Now I know I want to show that Z(G) is a subgroup iff for every a,b in Z(G), a*b^(-1) is in Z(G). I just don't understand how to do it.

Am I right to assume that x is in Z(G) also? (because it sure seems that way)

And I don't even know where to start with the last part. I think it's just asking me if Z(G) is abelian, which it is, but again, I have no clue what to do with this problem.

Thanks.

[Woozie]

I finished most of my review sheet for our exam but I just don't know where to start with this problem on subgroups. I'd appreciate if someone could point me in the right direction, and possible give me some tips on how to think about groups (since they seem to be my weak point).

14) Let G be a group and Z(G)= {z in G : z*x=x*z for all x in G}. Prove that Z(G) is a subgroup of G. Z(G) is called the center of G. If G is abelian, what is Z(G)?

Now I know I want to show that Z(G) is a subgroup iff for every a,b in Z(G), a*b^(-1) is in Z(G). I just don't understand how to do it.

Am I right to assume that x is in Z(G) also? (because it sure seems that way)

And I don't even know where to start with the last part. I think it's just asking me if Z(G) is abelian, which it is, but again, I have no clue what to do with this problem.

Thanks.

No, it's not asking you if Z(G) is abelian (the subgroup Z(G) is abelian. But that's not what they're asking). What the question is asking you to do is use the one step subgroup test to show that for any group, Z(G) is a subgroup of G).

To use this test, you first have to show that the set is non empty. Then you say "let a,b be any two elements of Z(G). Then <proof that ab^-1 is in G.

So to prove it's non empty, you just have to find one element in G that you know for sure is in Z(G). Let's look at the definition of Z(G)

Z(G)= {z in G : z*x=x*z for all x in G}

What this is saying is that if you commute with every element of the group, you're in Z(G). Do we know of any elements in G that commutes with every element in G?

Next, we assume we have two elements of G. We can name these elements a and b (or x and y, or whatever you want to call them). These elements may or may not be distinct, we really don't care which.

If we're assuming a and b are in Z(G), then we're basically assuming that a and b commute with all elements of Z(G). So we need to use this fact to prove that a*b^-1 is also in Z(G) (i.e. we need to prove that the product ab^-1 also commutes with everything in G).

Edit: My bad, is this one called the 1 step test or the 2 step test? It goes by one of those names but I can't remember.

Edit2: This is called the One Step test. I think I called it the two step test earlier. My bad, I don't usually think of these tests by name. I'll fix that.

[Woozie]

By the way, when using the subgroup tests, remember to ALWAYS show that the subgroup is non-empty first. When I took abstract algebra, sooooo many people lost so many points due to not showing that the set is non-empty.

[SDSD]

I was approaching it by the one-step test because I used it in an earlier proof, but I think the two-step test may be a bit easier. Thanks for pointing me in the right direction. I'm going to try and work on the one-step test proof now.

Pf// for the 2-step, we know e is in Z(G), so it's nonempty. Let a,b be in Z(G). Since Z(G) is a group, it also contains the inverses.

WTS that ab is in Z(G). Let z be any elt of G.

Then (ab)z
= a(bz) by associativity
= a(zb) b is in Z(G)
= (az)b by associativity
= (za)b a is in Z(G)
= z(ab) by associativity
Therefore, ab is in Z(G). Then by 2-step test Z(G) is a subgroup of G.

edit: if both are abelian, does that mean Z(G) and G are equal to each other?

[Woozie]

I'm going to try and work on the one-step test proof now.

Pf// for the 2-step

You said "I'm going to use the one step now" and then said "for the 2 step...". I'm assuming you meant to say "for the one step...".

we know e is in Z(G), so it's nonempty.

This is correct, but your professor may mark off for this. He may ask "How do you know that e is in Z(G)?". Of course, your answer to that would be "because ex = xe for any x in G". But even though it's obvious, you have to show this. So you should phrase it something like this:

"Since G is a group, it contains an identity element, e. Since ex = xe for any x in G, then e is an element of Z(G). Thus, Z(G) is non-empty"

Let a,b be in Z(G). Since Z(G) is a group, it also contains the inverses.

We don't yet know that Z(G) is actually a group. In fact, that's technically what we're trying to prove. So we can't make this statement. When it comes to the one step subgroup test, you don't actually have to show that inverses are in the group. I'll explain why in a moment.

WTS that ab is in Z(G). Let z be any elt of G.

Then (ab)z
= a(bz) by associativity
= a(zb) b is in Z(G)
= (az)b by associativity
= (za)b a is in Z(G)
= z(ab) by associativity
Therefore, ab is in Z(G). Then by 2-step test Z(G) is a subgroup of G.

edit: if both are abelian, does that mean Z(G) and G are equal to each other?

You're on the right track. For the one step, you need to show that if you take any two elements of our subset (which is Z(G)), then the first element times the inverse of the second element is in our original group.

So if we let a,b be elements of our subset Z(G), then we need to show that ab^-1 is also in Z(G). To show that ab^-1 is in Z(G), we have to apply the definition of Z(G). To be in Z(G), you must commute with every element in G,

So the way you should have phrased this part is

"Let a,b be any two elements of Z(G)" (That's pretty much always going to be your first sentence after proving non-empty when you're using the one step test.)

So anyways:

"Let a,b be any two elements of Z(G). Then I must show that ab^-1 is also an element of Z(G)"

To do this, you're going to want to apply the definition of Z(G). If we want to show that ab^-1, then we need to show that for any g in G, (ab^-1)g = g(ab^-1).

So your final proof will look something like this:

"Since G is a group, it contains an identity element, e. Since ex = xe for any x in G, then e is an element of Z(G). Thus, Z(G) is non-empty

Let a,b be any two elements of Z(G). Then I must show that ab^-1 is also an element of Z(G). To prove this, let g be any element of G. Then <proof that (ab^-1)g = g(ab^-1)>"

The proof that (ab^-1)g = g(ab^-1) is very similar to what you did above.

Earlier I said that you don't have to worry about proving that inverses are in Z(G). The reason for this is because if Z(G) is non-empty and ab^-1 is in Z(G) (for any a,b in Z(G)), then it follows that inverses are automatically in.

To see why this is true, suppose K is some subset of a group G. Suppose K passes the two step subgroup test. Let j be in element of K. Since K passes the two step test, we know that if we pick j and any other element (let's call it k), then jk^-1 is in K. So what if for our second element, we pick j again? So instead of picking j and k, we pick j twice. Then jj^-1 is in K. But jj^-1 is e. So K automatically contains e if it passes the two step test.

Now we pick two more elements. This time we pick e as our first and j as our second. Since K passes the two step test, ej^-1 is in K. But ej^-1 = j^-1. Thus, if K passes the two step test, all inverses are automatically in K. Thus, when we're applying the test, we don't even care about inverses because if the group passes the test, the inverses come automatically. If anything about my explanation of why we don't worry about inverses confuses you, look in your A.A. textbook and read their proof of the one step test.

It's a little weird that we call it the one step test. Technically there are two steps (showing non-empty and showing ab^-1 is in the subset).

Edit:

edit: if both are abelian, does that mean Z(G) and G are equal to each other?

No. To see why, consider the group Z_8 (integers under mod 8 addition). Then Z_8 is abelian. The set {0,2,4,6} under modulo 8 addition is a subgroup of Z_8 (I think. I've taken sleeping pills so I'm a little off right now. And by a little, I'm mean very). {0,2,4,6} is also abelian. But even though Z_8 is abelian and {0,2,4,6} is an abelian subgroup of Z_8, {0,2,4,6} does not equal Z_8.

[SDSD]

I think I have it now. I had to play around with it a little bit before I got it. I did get mixed up before and typed up the wrong work. And the identity explanation makes a lot of sense.

Since G is a group, it contains an identity element, e. Since ex = xe for any x in G, then e is an element of Z(G). Thus, Z(G) is non-empty

Let a,b be any two elements of Z(G). Then I must show that ab^-1 is also an element of Z(G). To prove this, let g be any element of G. Then <proof that (ab^-1)g = g(ab^-1)>

Let a,b be in Z(G). Choose some z in G. Then zb = bz. After multiplying by the inverse, z = bzb^-1, and multiplying again b^-1z = zb^-1. (obviously b^-1 is in Z(G))

Use this fact to show that ab^-1 is in Z(G).

(ab^-1)z
= a(b^-1z)
= a(zb^-1)
= (az)b^-1
= (za)b^-1
= z(ab^-1)

So ab^-1 is in Z(G).

I'm really positive this is right (I hope it is because I have analysis to do tonight). Thanks for all the help, much better than asking my TA for an explanation.

[willriker]

stupid number jumbling really, i dont see the point in this test. Is it just to show the associative property? If so, you really dont need to go through all that testing to learn this.

[Bobbity]

No. To see why, consider the group Z_8 (integers under mod 8 addition). Then Z_8 is abelian. The set {0,2,4,6} under modulo 8 addition is a subgroup of Z_8 (I think. I've taken sleeping pills so I'm a little off right now. And by a little, I'm mean very). {0,2,4,6} is also abelian. But even though Z_8 is abelian and {0,2,4,6} is an abelian subgroup of Z_8, {0,2,4,6} does not equal Z_8.

Just a quick clarification:

If G is abelian, then Z(G) is the entire group. While {0,2,4,6} is a subgroup of Z_8, it is not the center of the group (i.e., Z(G)). For example, 1+x=x+1 for every x in Z_8, so 1 is in the center (yet isn't in the subgroup you provided).

The center essentially picks out all associative elements of a group. If a group is already abelian, the subgroup of associative elements (i.e., the center) is simply the entire group.

[Woozie]

I think I have it now. I had to play around with it a little bit before I got it. I did get mixed up before and typed up the wrong work. And the identity explanation makes a lot of sense.



Let a,b be in Z(G). Choose some z in G. Then zb = bz. After multiplying by the inverse, z = bzb^-1, and multiplying again b^-1z = zb^-1. (obviously b^-1 is in Z(G))

Use this fact to show that ab^-1 is in Z(G).

(ab^-1)z
= a(b^-1z)
= a(zb^-1)
= (az)b^-1
= (za)b^-1
= z(ab^-1)

So ab^-1 is in Z(G).

I'm really positive this is right (I hope it is because I have analysis to do tonight). Thanks for all the help, much better than asking my TA for an explanation.

Your proof is correct.

Just a quick clarification:

If G is abelian, then Z(G) is the entire group. While {0,2,4,6} is a subgroup of Z_8, it is not the center of the group (i.e., Z(G)). For example, 1+x=x+1 for every x in Z_8, so 1 is in the center (yet isn't in the subgroup you provided).

The center essentially picks out all associative elements of a group. If a group is already abelian, the subgroup of associative elements (i.e., the center) is simply the entire group.

My bad, I thought he was asking if an Abelian subgroup of an Abelian group is the whole group. But now that I reread the question, he clearly wasn't asking what I thought he was. Like I said, sleeping pills... x_x