Asking for a check of my proof (group theory)

[Boi]

This is an exercise from Herstein's "Topics in Algebra":

The thing about this exercise is, I think that it is meant to combine from several previous exercises, so I will use results from these.
Proof
First things first, the results needed ("$H \le G$" means "$H$ is a subgroup of $G$" and $i_G(H)$ is the index of $H$ in $G$):

• (4a) if $H \le G$, then $gHg^{-1} \le G \quad (\forall g \in G)$.
• (12) if $H \le G$, $K \le G$, $i_G(H) < \infty$ and $i_G(K) < \infty$, then $i_G(H \cap K) \le i_G(H) \cdot i_G(K)$.
• (18) if $H \le G$ and $N= \bigcap\limits_{g \in G} gHg^{-1}$, then $N \le G$ and $gNg^{-1} = N \quad (\forall g \in G)$.
• (*19) if $H \le G$ and $i_G(H) < \infty$, then there is only a finite number of subgroups of the form $gHg^{-1} \quad (g \in G)$ and that their number is bounded above by $i_G(H)$.

Now I would like to show that $i_G(H)=i_G(gHg^{-1}) \quad (\forall g \in G)$. Proof: define sets $L=\{aH \; | \; a \in G\}$ and $L' = \{ (a)gHg^{-1} \; | \; a \in G \}$. Now I also define a function $f: L \mapsto L'$ as $f(aH)=(gag^{-1})gHg^{-1}$. Now I just show that $f$ is a well-defined bijection:

• well-definition: suppose $aH=bH$. Then, if I multiply by $g$ on the left and by $g^{-1}$ on the right I get $gaHg^{-1}=gbHg^{-1}$. But then, using associativity and inserting neutral element as $e=g^{-1}g$ I get $gaHg^{-1}=gbHg^{-1} \Rightarrow gaeHg^{-1}=gbeHg^{-1} \Rightarrow (gag^{-1})gHg^{-1}=(gag^{-1})gHg^{-1}$. So, $aH=bH \Rightarrow f(aH)=f(bH)$.
• injectivity: suppose $f(aH)=f(bH)$. So, $(gag^{-1})gHg^{-1}=(gbg^{-1})gHg^{-1} \Rightarrow gaHg^{-1}=gbHg^{-1}$. But then, by cancelling $g$ and $g^{-1}$, I have $aH=bH$. Hence, $f(aH)=f(bH) \Rightarrow aH=bH$.
• surjectivity: for every $(a)gHg^{-1} \in L'$ there is an element of $L$, namely $g^{-1}agH$, that maps to it. ($f(g^{-1}agH)=(gg^{-1}agg^{-1})gHg^{-1}=(eae)gHg^{-1}=agHg^{-1}$).

Because there is a bijection between $L$ and $L'$, they must have the same cardinality, giving me $i_G(H)=i_G(gHg^{-1})$.
Back to the exercise, I will use subgroup $N$ from 18. $N$ is clearly contained in $H$ since $H$ is in the set of groups whose interesection is taken in the definition of $N$ ($H=eHe^{-1}$). Then, combining the results of 4a, 12, *19 and the previously proved proposition, $i_G(N) \le i_G(H)^{i_G(H)}$ (I think the use of 4a and previous proposition is pretty clear. 12 and *19 give the actual bound, with 12 telling me to multiply the index of $H$ by itself and *19 ensuring that the number of such multiplications is finite.)
Thanks for checking. I, again, apologise if the thread is confusing.

 

[fresh_42]

Since you asked how we read proofs let me tell you how I approach this proof.

I start by trying to grasp the problem. We have a subgroup $H\leq G$ of finite index, i.e. $G=g_1H \cup \ldots \cup g_nH =H \overline{g}_1 \cup \ldots \cup H \overline{g}_n$ and we know from a previous problem that there are only finitely many distinct subgroups of the form $gHg^{-1}.$

We need to find a subgroup $N\leq H$ of finite index in $G$ that is a normal subgroup $N\trianglelefteq G,$ i.e. $G/N$ is a finite group.
In words: Every subgroup of finite index contains a normal subgroup that is of finite index, too.

If $H$ is finite, or $G$ is finite, we can choose $N=\{e\}.$ If $H \trianglelefteq G$ is normal, e.g. if $\left|G:H\right|=2,$ or $G$ is abelian, we can choose $N=H.$

This means, that we may assume that $H,G$ are infinite and that there is an element $g_0\in G$ such that $g_0Hg_0^{-1} \not\subseteq H.$

It is now clear that $N$ has to be small enough to be normal, and big enough to be of finite index.

These are my initial thoughts on the problem. I tried to get an idea of how to achieve this but haven't found one. The next step is: let's see what your idea was.

(To be continued ... (will take me some time))

 

[Boi]

Since you asked how we read proofs let me tell you how I approach this proof.

I start by trying to grasp the problem. We have a subgroup $H\leq G$ of finite index, i.e. $G=g_1H \cup \ldots \cup g_nH =H \overline{g}_1 \cup \ldots \cup H \overline{g}_n$ and we know from a previous problem that there are only finitely many distinct subgroups of the form $gHg^{-1}.$

We need to find a subgroup $N\leq H$ of finite index in $G$ that is a normal subgroup $N\trianglelefteq G,$ i.e. $G/N$ is a finite group.
In words: Every subgroup of finite index contains a normal subgroup that is of finite index, too.

If $H$ is finite, or $G$ is finite, we can choose $N=\{e\}.$ If $H \trianglelefteq G$ is normal, e.g. if $\left|G:H\right|=2,$ or $G$ is abelian, we can choose $N=H.$

This means, that we may assume that $H,G$ are infinite and that there is an element $g_0\in G$ such that $g_0Hg_0^{-1} \not\subseteq H.$

It is now clear that $N$ has to be small enough to be normal, and big enough to be of finite index.

These are my initial thoughts on the problem. I tried to get an idea of how to achieve this but haven't found one. The next step is: let's see what your idea was.

(To be continued ... (will take me some time))

Just want to quickly mention that the notions of normal subgroups and quotient groups have not been introduced in the book yet.

 

[fresh_42]

Just want to quickly mention that the notions of normal subgroups and quotient groups have not been introduced in the book yet.

Normality simply means $N_G(N)=\{a\in G\,|\,aNa^{-1}=N\}=G$ and normal or $N\trianglelefteq G$ was shorter to write than the formula.

Btw. the notation I know is $i_G(H)=\left|G:H\right| .$ Here is a link for condition (12):

Index of a subgroup - Wikipedia

en.wikipedia.org

(4a) and (18) are easy to see and (19) was:

Could somebody help me refine my proof, please? (group theory)

This is an exercise from Herstein's "Topics in algebra":G What follows isn't really a proof, more so a collection of my main ideas, which I believe should lead to a proof. I kindly ask anyone reading to not only check whether these ideas are correct, but also point out which ones ought to be...

www.freemathhelp.com

Remark #1: Your proof of $i_G(H)=i_G\left(gHg^{-1}\right)$ is a bit too complicated. The finite index means that we can write
$G=g_1\cdot H\,\cup\,\ldots\,\cup\, g_n\cdot H$ and conjugation with an element $g \in G$ does not change this.

 

[fresh_42]

I will use subgroup $N$ from 18. $N$ is clearly contained in $H$ since $H$ is in the set of groups whose interesection is taken in the definition of $N$ ($H=eHe^{-1}$).

Nice trick! That's it. It guarantees normality and intersecting with $eHe^{-1}$ was the clue I have been looking for!

Let's look at the index. I have some trouble understanding your argument, probably because I find formulas easier to read - despite the fact that I am a great fan of van der Waerden's algebra books. I'm also a bit hesitant if someone speaks about cardinalities if infinite sets are involved. That requires caution! So let me write it down.
$$\begin{array}{lll} i_G(N)&=\displaystyle{i_G\left(\bigcap_{g\in G}gHg^{-1}\right)\quad \text{finite intersection by (19)}}\\[18pt] &=\displaystyle{i_G\left(\bigcap_{k=1}^n \,g_kHg_k^{-1}\right)\quad \text{for some }g_k \in G}\\[18pt] &\leq \displaystyle{\prod_{k=1}^n i_G\left(g_kHg_k^{-1}\right)}\quad \text{by (12)}\\[18pt] &=n\cdot i_G(H)\quad \text{by the previous consideration}\\[18pt] &< \infty \quad \text{by the finite index of }H \leq G \end{array}$$
Last remark: We don't have an upper bound for $n,$ do we? I mean except for the cardinality of the group of inner automorphisms of $G$ which could be infinite.

 

[Boi]

$$\begin{array}{lll} \displaystyle{\prod_{k=1}^n i_G\left(g_kHg_k^{-1}\right)}\quad \text{by (12)}\\[18pt] &=n\cdot i_G(H)\quad \text{by the previous consideration}\\[18pt] \end{array}$$
Last remark: We don't have an upper bound for $n,$ do we? I mean except for the cardinality of the group of inner automorphisms of $G$ which could be infinite.

It's a product, so you must have meant $i_G(H)^{n}$.

Last remark: We don't have an upper bound for $n,$ do we? I mean except for the cardinality of the group of inner automorphisms of $G$ which could be infinite.

No, I think we do have a nice upper bound. In my proof of *19 (which I thank you for helping me refine) I have shown that the set of left cosets of a subgroup $H$ maps onto the set of conjugates of $H$.

 

[fresh_42]

It's a product, so you must have meant $i_G(H)^{n}$.

Oops!

No, I think we do have a nice upper bound. In my proof of *19 (which I thank you for helping me refine) I have shown that the set of left cosets of a subgroup $H$ maps onto the set of conjugates of $H$.

You are right! I first thought that $i_G(H)^{i_G(H)}$ was a typo. (I think I need sleep. Obviously.)