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Permutation representation argument validity

kalish

Member
Oct 7, 2013
99
Hello,
I would like to check if the work I have done for this problem is valid and accurate. Any input would be appreciated. Thank you.

**Problem statement:** Let $G$ be a group of order 150. Let $H$ be a subgroup of $G$ of order 25. Consider the action of $G$ on $G/H$ by left multiplication: $g*aH=gaH.$ Use the permutation representation of the action to show that $G$ is not simple.

**My attempt:** Let $S_6$ be the group of permutations on $G/H$. Then, the action of $G$ on $G/H$ defines a homomorphism $f:G \rightarrow S_6$. We know $|S_6| = 720.$ Since $|G|=150$ does not divide 720, and $f(G)$ is a subgroup of $S_6$, $f$ cannot be one-to-one. Thus, $\exists$ $g_1,g_2$ distinct in $G$ such that $f(g_1)=f(g_2) \implies f(g_1g_2^{-1})=e$. Thus, $\ker(f) = \{g:f(g)=e\}$. Since $\ker(f)$ is a normal subgroup of $G$, we have found a normal subgroup of $G$. Also, since $f$ is non-trivial, then $\ker(f)$ is a proper normal subgroup of $G.$ Hence $G$ is not simple.

Any suggestions or corrections?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
I think it is clear that since $|G|$ does not divide 720, $\text{ker}(f)$ is a non-trivial normal subgroup of $G$, so there is no need to talk about the existence of $g_1,g_2$ or restate the definition of $\text{ker}(f)$.

I *do* think you should say WHY $f$ is not the trivial homomorphism. It's pretty simple, though:

Since $|H| < |G|$, we can take any $g \in G - H$, which takes (under the action) the coset $H$ to $gH \neq H$, so $f(G)$ contains at least one non-identity element: namely, $f(g)$.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
This is actually a special case of a theorem proved in Herstein, which goes as follows:

If $G$ is a finite group with a subgroup $H$ such that $|G| \not\mid ([G:H])!$ then $G$ contains a non-trivial proper normal subgroup containing $H$.

One obvious corollary is then that such a group $G$ cannot be simple.

You would be far better off adapting your proof to this more general one, which can be re-used in many more situations.
 

kalish

Member
Oct 7, 2013
99
Hi,
Which Herstein book is this from? I would like to explore further.

Thanks.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
His classic Topics In Algebra​.
 

kalish

Member
Oct 7, 2013
99
That's what I found from my search as well. Do you have a copy of the book or know where I can find one?
 

kalish

Member
Oct 7, 2013
99
That sounds like a fantastic result. I cannot find the book anywhere though. Could you please reproduce the proof for me here, so that I could use it to study? I would really appreciate it.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Theorem 2.G (p. 62, chapter 2):


If $G$ is a group, $H$ a subgroup of $G$, and $S$ is the set of all right cosets of $H$ in $G$, then there is a homomorphism $\theta$ of $G$ into $A(S)$, and the kernel of $\theta$ is the largest normal subgroup of $G$ which is contained in $H$.

(a few words about notation: Herstein uses $A(S)$ to stand for the group of all bijections on $S$...if $|S| = n$, then $A(S)$ is isomorphic to $S_n$. Herstein also writes his mappings on the RIGHT, as in $(x)\sigma$ instead of $\sigma(x)$, so that composition and multiplication are "in the same order", instead of reversed. For this reason, he uses right cosets and right-multiplication instead of the left cosets (and left-multiplication) one often sees used in other texts. He also denotes the index of $H$ in $G$ by $i(H)$ , instead of $[G:H]$ and denotes $|G|$ by $o(G)$).

Proof: Let $G$ be a group, $H$ a subgroup of $G$. Let $S$ be the set whose elements are right cosets of $H$ in $G$. That is, $S = \{Hg: g \in G\}$. $S$ need not be a group itself, in fact, it would be a group only if $H$ were a normal subgroup of $G$. However, we can make our group $G$ act on $S$ in the following natural way: for $g \in G$ let $t_g:S \to S$ be defined by: $(Hx)t_g = Hxg$. Emulating the proof of Theorem 2.f we can easily prove:

(1) $t_g \in A(S)$ for every $g \in G$
(2) $t_{gh} = t_gt_h$.

Thus the mapping $\theta: G \to A(S)$ defined by $\theta(g) = t_g$ is a homomorphism of $G$ into $A(S)$. Can one always say that $\theta$ is an isomorphism? Suppose that $K$ is the kernel of $\theta$. If $g_0 \in K$, then $\theta(g_0) = t_{g_0}$ is the identity map on $S$, so that for every $X \in S, Xt_{g_0} = X$. Since every element of $S$ is a right coset of $H$ in $G$, we must have that $Hat_{g_0} = Ha$ for every $a \in G$, and using the definition of $t_{g_0}$, namely, $Hat_{g_0} = Hag_0$, we arrive at the identity $Hag_0 = Ha$ for every $a \in G$. On the other hand if $b \in G$ is such that $Hxb = Hx$ for every $x \in G$, retracing our argument we could show that $b \in K$. Thus $K = \{b \in G|Hxb = Hx$ all $x \in G\}$. We claim that from this characterization of $K,\ K$ must be the largest normal subgroup of $G$ which is contained in $H$. We first explain the use of the word largest; by this we mean if $N$ is a normal subgroup of $G$ which is contained in $H$, then $N$ must be contained in $K$. We wish to show this is the case. That $K$ is a normal subgroup of $G$ follows from the fact that it is the kernel of a homomorphism of $G$. Now we assert that $K \subset H$, for if $b \in K, Hab = Ha$ for every $a \in G$, so in particular, $Hb = Heb = He = H$, whence $b \in H$. Finally, if $N$ is a normal subgroup of $G$ which is contained in $H$, if $n \in N,\ a \in G$, then $ana^{-1} \in N \subset H$, so that $Hana^{-1} = H$; thus $Han = Ha$ for all $a \in G$. Therefore, $n \in K$ by our characterization of $K$.

**********

Remarks following the proof:

The case $H = (e)$ just yields Cayley's Theorem (Theorem 2.f). If $H$ should happen to have no normal subgroup of $G$, other than $(e)$ in it, then $\theta$ must be an isomorphism of $G$ into $A(S)$....(some text omitted)....

We examine these remarks a little more closely. Suppose that $G$ has a subgroup $H$ whose index $i(H)$ (that is, the number of right cosets of $H$ in $G$) satisfies $i(H)! < o(G)$. Let $S$ be the set of all right cosets of $H$ in $G$. The mapping, $\theta$, of Theorem 2.g cannot be an isomorphism, for if it were, $\theta(G)$ would have $o(G)$ elements and yet would be a subgroup of $A(S)$ which has $i(H)! < o(G)$ elements. Therefore, the kernel of $\theta$ must be larger than $(e)$; this kernel being the largest normal subgroup of $G$ which is contained in $H$, we can conclude that $H$ contains a nontrivial normal subgroup of $G$.

However, the above argument has implications even when $i(H)!$ is not less than $o(G)$. If $o(G)$ does not divide $i(H)!$ then by invoking Lagrange's theorem we know that $A(S)$ can have no subgroup of order $o(G)$, hence no subgroup isomorphic to $G$. However $A(S)$ does contain $\theta(G)$, whence $\theta(G)$ cannot be isomorphic to $G$, that is, $\theta$ cannot be an isomorphism. But then, as above, $H$ must contain a nontrivial normal subgroup of $G$. We summarize this as:

Lemma 2.21 If $G$ is a finite group, and $H \neq G$ is a subgroup of $G$ such that $o(G) \not\mid i(H)!$, then $H$ must contain a nontrivial normal subgroup of $G$. In particular, $G$ cannot be simple.

(Note to the moderating staff: although this is an excerpt from a copyrighted work, I believe this sample falls under the province of "Fair Use" for the purpose of "Scholarly research and exposition", and is not intended for commercial gain or to circumvent existing copyright laws).
 
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