Group theory

**signaldoc** · October 20th 2009, 11:01 AM

I am trying to do all the exercises in Rotman (3rd ed.). It's an integral part of his exposition to do many of them, but this one is optional. I don't get it... (problem 2.83 in the book).

G=H x K, N is normal in G. Prove: Either N is abelian OR N intersects one of the factors non-trivially.

Suppose N intersects H in the identity AND N intersects K in the identity (i.e., trivially in both cases). The proposition is then that N must be abelian.

I think we have the following fact: Every element of N must be of the form n=hk with neither h,k the identity unless both are. From this, we can deduce that every first component, h, has the same order, m, as every second component, k.

h^m=1
k^m=1

But does this make N abelian? No.

Note: Presumably, Sylow theory is not used to solve this, since it precedes that theory.

Any help will be greatly appreciated, and apologies in advance if I'm being dense.

**Opalg** · October 20th 2009, 12:34 PM

Suppose that N intersects each of H and K trivially. Let $n = h_0k_0\in N$ , and let $h_1\in H$ . Then $n^{-1} = h_0^{-1}k_0^{-1}\in N$ . Also, $h_1^{-1}h_0h_1k_0\in N$ (by normality). Therefore the product $h_0^{-1}k_0^{-1}h_1^{-1}h_0h_1k_0 = h_0^{-1}h_1^{-1}h_0h_1$ is in N. But N intersects H trivially, so $h_0^{-1}h_1^{-1}h_0h_1$ is the identity. Thus $h_0$ is in the centre of H. Similarly $k_0$ is in the centre of K. That's enough to show that N is abelian.

**tonio** · October 20th 2009, 12:55 PM

Originally Posted by signaldoc

I am trying to do all the exercises in Rotman (3rd ed.). It's an integral part of his exposition to do many of them, but this one is optional. I don't get it... (problem 2.83 in the book).

G=H x K, N is normal in G. Prove: Either N is abelian OR N intersects one of the factors non-trivially.

Suppose N intersects H in the identity AND N intersects K in the identity (i.e., trivially in both cases). The proposition is then that N must be abelian.

I think we have the following fact: Every element of N must be of the form n=hk with neither h,k the identity unless both are. From this, we can deduce that every first component, h, has the same order, m, as every second component, k.

h^m=1
k^m=1

But does this make N abelian? No.

Note: Presumably, Sylow theory is not used to solve this, since it precedes that theory.

Any help will be greatly appreciated, and apologies in advance if I'm being dense.

Ok, what you did was fine: take $n_1\,,\,n_2\in N\Longrightarrow\,n_1=h_1k_1\,,\,\,n_2=h_2k_2\,,\, \,with\,\,\,h_i\in H\,\,and\,\,k_i\in K$

Now use that both $H,K$ are normal subgroups to show that:

$[h,n]=[k,n]=1\,\,\forall h\in H ,\,\,n\in N$

Finally, use the basic relation between commutators $[xy,z]=[x,z]^y[y,z]$ , so that finally: $[n_1,n_2]=[h_1k_1,n_2]\Longleftrightarrow N$ is abelian.

Tonio

**signaldoc** · October 20th 2009, 12:58 PM

Very quick....you must do this in your sleep. I guess that's why I need to do the problems... Thanks!

**signaldoc** · October 20th 2009, 01:04 PM

Originally Posted by tonio

Ok, what you did was fine: take $n_1\,,\,n_2\in N\Longrightarrow\,n_1=h_1k_1\,,\,\,n_2=h_2k_2\,,\, \,with\,\,\,h_i\in H\,\,and\,\,k_i\in K$

Now use that both $H,K$ are normal subgroups to show that:

$[h,n]=[k,n]=1\,\,\forall h\in H ,\,\,n\in N$

Finally, use the basic relation between commutators $[xy,z]=[x,z]^y[y,z]$ , so that finally: $[n_1,n_2]=[h_1k_1,n_2]\Longleftrightarrow N$ is abelian.

Tonio

Much thanks!

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Thanks... you guys are good!

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