Obtaining a larger group from a given group

**kalyanram** · March 12th 2011, 10:51 AM

[Excerpt from Topics in Abstract Algebra Second Edition by I.N.Herstein pg 69]

This is what Herstein talks of verbatim:

Generally, if $G$ is a group, $T$ an automorphism of order $r$ of $G$ which is not an inner automorphism, pick a symbol $x$ and consider all elements $x^ig$ , $i= 0, \pm1, \pm2,.... g \in G$ subject to $x^ig = x^i^'$ $g^'$ if and only if $i \equiv i^'$ $mod r$ , $g = g^'$ and $x^{-1}g^{i}x = gT^{i} \forall i$ . This way we obtain a larger group $\{G,T\}$ and $\{G,T\}/G \approx$ group generated by $T =$ cyclic group of order $r$ .

I have the following questions
1. What is the nature of the "a larger group $\{G,T\}$ " under discussion I mean the nature of the elements the operation.
2. As I understand this symbol $x$ that Herestein talks of abides to the binary operation of $G$ . Correct?
3. Is it correct to assume that $x^r = e$ where $e$ is the identity element.
4. I have difficulty imagining $G$ in $\{G,T\}$ but I guess I will come to that once I am clear on the behavior of $\{G,T\}$ as a group.

Thanks in advance.

~Kalyan.

**Tinyboss** · March 12th 2011, 03:48 PM

Does $gT^i$ mean the image of g under i iterations of T? If so, are you sure it's not supposed to be $x^{-i}gx^i=gT^i$ ?

**kalyanram** · March 12th 2011, 07:22 PM

Hi Tiny,
Yes $gT^i$ stands for applying $T$ iteratively $i$ times on $g$ . I do agree that this notation is confusing. In fact if we take $i=1$ we end up with $x^{-1}gx = gT$ and on iteratively applying $T-i$ times on g we do get $x^{-i}gx^{i} = gT^i$ .

Here is a note I tried checking this section in Topics in Abstract Algebra by I.N.Herestein 3rd edition (edited by Barbara Cortzen and David J.Winter) and its omitted entirely. The preface does mention "certain typographical errors and notation inconsistencies in 2nd edition" probably it could be due to that. However the paragraph I have typed in is exactly same as it appears in the 2nd edition.

**Opalg** · March 13th 2011, 04:19 AM

Originally Posted by kalyanram

[Excerpt from Topics in Abstract Algebra Second Edition by I.N.Herstein pg 69]

This is what Herstein talks of verbatim:

Generally, if $G$ is a group, $T$ an automorphism of order $r$ of $G$ which is not an inner automorphism, pick a symbol $x$ and consider all elements $x^ig$ , $i= 0, \pm1, \pm2,.... g \in G$ subject to $x^ig = x^i^'$ $g^'$ if and only if $i \equiv i^'$ $mod r$ , $g = g^'$ and $x^{-1}g^{i}x = gT^{i} \forall i$ . This way we obtain a larger group $\{G,T\}$ and $\{G,T\}/G \approx$ group generated by $T =$ cyclic group of order $r$ .

I have the following questions
1. What is the nature of the "a larger group $\{G,T\}$ " under discussion I mean the nature of the elements the operation.
2. As I understand this symbol $x$ that Herestein talks of abides to the binary operation of $G$ . Correct?
3. Is it correct to assume that $x^r = e$ where $e$ is the identity element.
4. I have difficulty imagining $G$ in $\{G,T\}$ but I guess I will come to that once I am clear on the behavior of $\{G,T\}$ as a group.

This is a description of one way to construct the semidirect product $G\rtimes_T\mathbb{Z}_r$ of G by the action of $\mathbb{Z}_r$ induced by the automorphism T.

A more concrete way to make this construction is to say that $G\rtimes_T\mathbb{Z}_r$ consists of pairs $(g,i)\in G\times \mathbb{Z}_r$ , with the group operation defined by $\boxed{(g,i)*(h,j) = ((gT^j)h,i\mathbin{\dot+}j)}$ (where the dotted plus means addition mod r).

In this construction, you can identify G with the (normal) subgroup $\{(g,0): g\in G\}$ . The connection with Herstein's construction is that $x^i$ corresponds to the element $(e,i)$ , where e is the identity element of G.

The purpose of this construction is that given the group G and the (outer) automorphism T, it embeds G in the larger group $G\rtimes_T\mathbb{Z}_r$ , in which the automorphism becomes inner (because $(gT,0) = (e,-1)(g,0)(e,1)$ , or in Herstein's notation $gT = x^{-1}gx$ ).

**kalyanram** · March 14th 2011, 11:32 AM

Hi Oplag,
Thanks for the reply it has at least opened a new dimension to think on. I am not able to grasp what you said completely as of now as I have a little understanding of direct products and semi direct product of a group. I will ping you again once I am clear on this.

Kalyan.

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