Proving $H\subset gHg^{-1}$ without the normality condition

Question

In most textbooks a subgroup $H$ of group $G$ is defined as normal if $$i) \quad gHg^{-1} \subset H \quad \forall g \in G.$$ I understand that they don't write $gHg^{-1} = H$ as the definition of normality because we can deduce from condition "i" that $H \subset gHg^{-1}$. My question is if $H$ is a subgroup of $G$ is it true in general that $H \subset gHg^{-1}$?

Answer 1

No, but before I provide a counterexample, note that the map $\gamma_g=a\mapsto gag^{-1}$ is a bijection at least, since it has an inverse in $\gamma_{g^{-1}}=a\mapsto g^{-1}ag$. If $H$ is finite, this would trivially imply that $H=gHg^{-1}$ and $H$ would therefore be normal. I do not know if you have considered that not every subgroup of a finite group is normal. But here is an example, in case not.

In $S_4$, we may consider the cyclic subgroup of order $2$, $\langle(12)(34)\rangle$. The only non-identity element is $(12)(34)$ and if we conjugate this element by $(123)$ we get $$(123)*(12)(34)*(132)=(13)(24)$$ $\langle(13)(24)\rangle$ also has order $2$, and in particular, does not contain $(12)(34)$ as an element, so $$\langle(12)(34)\rangle\not\subseteq(123)\langle(12)(34)\rangle(123)^{-1}$$

Answer 2

We have the following two conditions:

1. $g H g^{-1} ⊆ H$ for every $g ∈ G$,

2. $H ⊆ g H g^{-1}$ for every $g ∈ G$.

You have already stated that condition 2 can be deduced from condition 1. But in the same way, condition 1 can also be deduced from condition 2! In other words, conditions 1 and 2 are actually equivalent. You are therefore asking if every subgroup is normal, which is not the case.

Answer 3

Let $G$ be a group, $H$ a subgroup.

1. For a fixed $g\in G$, $gHg^{-1}\subseteq H\iff H\subseteq g^{-1}Hg$.

Indeed, from $gHg^{-1}\subseteq H$, multiply by $g^{-1}$ on the left and $g$ on the right to get $H\subseteq g^{-1}Hg$; from $H\subseteq g^{-1}Hg$, multiply on the left by $g$ and the right by $g^{-1}$ to get $gHg^{-1}\subseteq H$.

2. For all $g\in G$, $gHg^{-1}\subseteq H \iff$ for all $g\in G$, $H\subseteq gHg^{-1} \iff$ for all $g\in G$, $gHg^{-1}=H$.

If $gHg^{-1}\subseteq H$ for all $g\in G$, then it holds for $g^{-1}$: $g^{-1}H(g^{-1})^{-1}=g^{-1}Hg\subseteq H$. That implies $H\subseteq gHg^{-1}$, by 1. Similarly, if $H\subseteq gHg^{-1}$ holds for all $g$, then applying 1 to $H\subseteq g^{-1}Hg$ gives $gHg^{-1}\subseteq H$. The third equivalence is then clear.

3. If $H$ is finite, then for a fixed $g$, $gHg^{-1}\subseteq H \iff H\subseteq gHg^{-1}\iff gHg^{-1}=H$. For infinite $H$, this need not hold.

From $gHg^{-1}\subseteq H$, by 1 we get $H\subseteq g^{-1}Hg$. If $H$ is finite, then since $g^{-1}Hg$ has the same number of elements as $H$, we therefore have $H=g^{-1}Hg$. This gives $g^{-1}Hg\subseteq H$, and applying 1 again we get $H\subseteq gHg^{-1}$. The converse follows similarly.

For $H$ infinite, an example where $gHg^{-1}\subseteq H$ but $H\not\subseteq gHg^{-1}$ can be found here.

If we define "a subgroup $H$ is normal in $G$" as "for all $g\in G$, $gHg^{-1}=H$, we get the following equivalences, which are left as an exercise for the reader:

Theorem. Let $G$ be a group, and let $H$ be a subgroup of $G$. Assume $X\subseteq G$ and $Y\subseteq H$ are subgroups such that $\langle X\rangle = G$ and $\langle Y\rangle = H$. The following are equivalent

1. $H\triangleleft G$.
2. For all $g\in G$, $gHg^{-1}\subseteq H$.
3. For all $g\in G$, $H\subseteq gHg^{-1}$.
4. For all $g\in G$, for all $h\in H$ $ghg^{-1}\in H$.
5. For all $g\in G$, for all $y\in Y$, $gyg^{-1}\in H$.
6. For all $x\in X$, $xHx^{-1}=H$.
7. For all $x\in X$, $xHx^{-1}\subseteq H$ and $x^{-1}Hx\subseteq H$.
8. For all $x\in X$, for all $y\in Y$, $xyx^{-1}\in H$ and $x^{-1}yx\in H$.