1. ## Subgroup/Coset problem

Herstein ask us to find a group G, H<G and g in G such that gHg* is in H but gHg*/=H.
I can't imagine that this is true (of course if Herstein says it is true then he is correct) because of the following argument.
o(gH)=o(H) since gH has no duplications because G has the cancellation property. Now gH has o(H) distinct elements of G. Then o(gHg*)=o(H) again because of the cancellation law in G. Does it not follower that gHg* = H??? Where am I going wrong in my argument?
Thanks for any help offered,
Jomo

2. ## Re: Subgroup/Coset problem

It may seem plausible to you that since $gHg^{-1}$ and $H$ share many of the same properties, they must be the same. However, I assure you this is NOT the case. Let's explore this a little bit, before we finally answer the question.

Suppose $G$ is abelian. Then $Hg = \{hg: h \in H\} = \{gh: h \in H\} = gH$, since for every $g,h \in G,\ gh = hg$.

It follows, then, that $gHg^{-1} = \{ghg^{-1}:h \in H\} = \{gg^{-1}h: h \in H\} = \{h: h\in H\} = H$. So if we want to have ANY hope of finding a counter-example, we need to look in a non-abelian group.

The smallest non-abelian group is of order 6, and can be written like so:

$G = \{e,a,a^2,b,ab,a^2b\}$

where we evaluate products using the 3 rules:

1. $a^3 = e$
2. $b^2 = e$
3. $ba = a^2b$ <---this rule means $G$ cannot be abelian.

For example: $(a^2b)(ab) = a^2(ba)b = a^2(a^2b)b = a^4b^2 = a^4e = a^4 = a(a^3) = ae = a$.

You may wonder whether or not this really is a group, because although we have an identity, I have not proven the product is associative, or that we even have inverses. That is a valid question, and the best defense I can offer (without writing for 3 or 4 pages) is that you think of $a$ as the mapping of the plane which rotates it counter-clockwise 120 degrees, and think of $b$ as reflection about the $x$-axis. With a little thought, you might be able to come up with 2x2 matrices that perform these tasks, and matrix multiplication is associative, and closure can be shown by showing these 6 matrices all preserve an equilateral triangle (we have 3 rotations, 3 reflections). All of these matrices are invertible. It's fun to do, you should try proving this.

But, to this matter at hand: we need to find a suitable subgroup that "behaves badly". Let's look at $H = \langle b\rangle$. Since $b \neq e$, and $b^2 = e$, we see that $b$ is of order 2, so $H = \langle b\rangle = \{e,b\}$.

Now let's look at $aHa^{-1}$.

Since $a^3 = e$, we have that: $a(a^2) = e$, so that evidently, $a^{-1} = a^2$. So we need to evaluate two products: $aea^2$ and $aba^2$ (since we have two elements in $H$).

Now $aea^2 = aa^2 = e$, so far, so good.

$aba^2 = a(ba)a = a(a^2b)a = (a^3)(ba) = ba = a^2b$. Huh.

So for THIS subgroup $H$ of this group $G$, we have $aHa^{-1} = \{e,a^2b\} \neq \{e,b\} = H$.

Note that $(a^2b)^2 = (a^2b)(a^2b) = (a^2)(ba)(ab) = (a^2)(a^2b)(ab) = a^4(bab) = a(bab) = a(ba)b = a(a^2b)b = a^3b^2 = ee = e$.

So $a^2b$ is ALSO an element of order 2, and this might help explain WHY this occurs: we have two "copies" of the same subgroup in $G$, and conjugating by $a$ sends one to the other (there is actually a third copy:

$a^2Ha = \{e,ab\}$ so conjugating by $a$ (if you don't know already: conjugating by $a$ is the mapping $g \mapsto aga^{-1}$, for any $g \in G$) sends these 3 subgroups in a 3-cycle:

$H \to aHa^2 \to a^2Ha \to H$).

So Herstein is correct, but fair warning: math texts CAN contain typographical errors, and "proof by authority" is bad form for a mathematician, you are encouraged to use logic instead .

3. ## Re: Subgroup/Coset problem

Deveno,
Thank you for your excellent explanation. I understood it all including what you said about the equilateral triangle (I seen that group before). I will look over the last part so I understand it at a higher level.
I still do not understand why what I wrote was wrong. I thought that if set A is in set B and they had the same cardinality then A=B. I was given that gHg* is in H and showed(?) that o(gHg*) = o(H). Does not that mean that gHg*=H. Now since you showed me a counter example I KNOW that my argument is not valid but can you please tell me where it falls apart. BTW I did know that a counter example had to come from a non-abelian group. I was at least not that lost!
Thank you very much,
Steven

4. ## Re: Subgroup/Coset problem

Deveno,
Although i truly enjoyed reading what you wrote I now realized it is not a valid counterexample. Herstein stated to find a group G, a subgroup H of G and a g in G st gHg* IS IN H but gHg* is not H. I will re-read what you wrote and see if I can find a 'g' in G that works (but I doubt it based on my previous argument--Maybe Herstein is wrong!). Please respond as I am sure that I will enjoy reading your response.
Thanks,
Jomo

5. ## Re: Subgroup/Coset problem

I computed gHg* for all g in G down to the element level and all gHg* had two elements as expected. For some g, gHg*=H and for others gHg*/= H.
Please understand that Herstein is a giant and for me, a community college instructor with a strong masters in math from CCNY, to challenge him...well that is not my style. But this time I guess he is wrong and I am right!
Jomo

7. ## Re: Subgroup/Coset problem

One can show that the mapping:

$\phi: \Bbb Z \to \text{GL}_2(\Bbb Q)$ given by:

$\phi(k) = \begin{bmatrix}1&k\\0&1 \end{bmatrix}$

is an injective group homomorphism.

If we set, $H = \phi(\Bbb Z)$, and $a = \begin{bmatrix}2&0\\0&1 \end{bmatrix}$, what is $aHa^{-1}$?

(My apologies for mis-reading the original post).