# Prove that a Subgroup is Normal

I'm trying to understand normal subgroups and kernels of homomorphisms. Normal subgroups are defined as such:

$gHg^{-1}=H~~~\forall g \in G$

While i was trying to see which subgroups are normal, to verify the above statment, i should run through all elements of $G$.

To find a shorter way, I come up with the below:

$hGh^{-1}\cap H=\emptyset~~~\forall h \in H$ (wrong)

$h(G \setminus H)h^{-1}\cap H=\emptyset~~~\forall h \in H$ (correct)

This is basically telling that, if H is normal, then $hGh^{-1}$ is in $G \setminus H$.

Assume the contrary,

$h_{0}Gh_{0}^{-1}=h_{1}$ implies $h_{0}G=h_{3}$ implies $G=h_{4}$ or $G=H$ which is a contradiction.

Could anybody prove the above statement from the first, or possible vice versa?

Second one cheaper since H has fewer elements to run through, so i would prefer to write tests over the second one.

Regards.

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There are other ways to check whether a given subgroup $H$ is normal in a given group $G$. You can check whether the left cosets of $H$ in $G$ are also right cosets. You can try to find a homomorphism from $G$ to some other group $K$ with kernel $H$. If you already know the conjugacy classes of $G$ you can check whether $H$ is a union of conjugacy classes. – Gerry Myerson Mar 5 '12 at 22:39
The statement $h(G \setminus H)h^{-1} \cap H = \phi$ for all $h \in H$ is true for all subgroups $H$ of $G$, not just normal subgroups. – Derek Holt Mar 5 '12 at 23:44
... and you can just check $gHg^{-1}=H$ where $g$ runs over a set of generators of $G$ (you don't need to check every element of $G$). – Arturo Magidin Mar 6 '12 at 4:08

You want to show that $H$ is normal if and only if $h(G\setminus H)h^{-1} = G\setminus H$ for every $h\in H$.

This is incorrect. The condition $h(G\setminus H)h^{-1}=G\setminus H$ for all $h\in H$ holds whenever $H$ is a subgroup, whether or not it is normal: because the map $x\longmapsto hxh^{-1}$ is a bijection from $G$ to itself for every elements of $G$ (if $hxh^{-1}=hyh^{-1}$, then cancelling $h$ on the left and $h^{-1}$ on the right we get $x=y$; and for every $z\in G$, $z$ is the image of $h^{-1}zh$ under this map). If $H$ is a subgroup, and $h\in H$, then $hyh^{-1}\in H$ for every $y\in H$; that is, $hHh^{-1}=H$; and since we have a bijection and it maps $H$ to itself, it must map the complement of $H$ to itself, so $h(G\setminus H)h^{-1} = G\setminus H$. This does not require $H$ to be normal.

In fact, the condition may hold for sets that are not even subgroups. For instance, if $h\neq\{e\}$ then $X=\{h,h^{-1}\}$ is not a subgroup, but $hXh^{-1} = h^{-1}Xh = X$, so $X$ satisfies your condition.

So the condition does not characterize normality; it does not even characterize subgroup.

If you want a "cheaper" way to test normality, there are plenty: if you can find a homomorphism that has $H$ as a kernel, then $H$ is necessarily normal. In fact,

Let $G$ be a group. A subgroup $H$ of $G$ is normal in $G$ if and only if there exists a homomorphism $f$ with domain $G$ such that $\mathrm{ker}(f) = H$.

Proof. Kernels are normal; if $H$ is normal, then the canonical map $\pi\colon G\to G/H$ given by $\pi(g) = gH$ is a homomorphism, and $\mathrm{ker}(\pi)=H$. $\Box$

Another useful set of facts:

Let $G$ be a group, $H$ a subgroup, and $S$ a subset of $G$ such that $G=\langle S\rangle$. The following are equivalent:

1. $H$ is normal in $G$.
2. $gHg^{-1} = H$ for every $g\in G$.
3. $gHg^{-1}\subseteq H$ for every $g\in G$.
4. $sHs^{-1} = H$ for every $s\in S$.
5. $sHs^{-1}\subseteq H$ and $s^{-1}Hs\subseteq H$ for every $s\in S$.

If $H$ is finite, then in 5 it suffices to check $sHs^{-1}$ for every $s\in S$.

Proof. 1 and 2 are equivalent by definition. 2 clearly implies 3; if 3 holds and $x\in G$, then $xHx^{-1}\subseteq H$; we only need to show that $H\subseteq xHx^{-1}$ also holds; by 2, $(x^{-1})H(x^{-1})^{-1} = x^{-1}Hx\subseteq H$; multiplying on the left by $x$ and on the right by $x^{-1}$ we obtain $H = xx^{-1}Hxx^{-1}\subseteq xHx^{-1}$. This proves $xHx^{-1}=H$ for every $x\in G$, which means 2 holds.

2 implies 4; to see that 4 implies 3, first note that since $sHs^{-1}=H$ for every $s\in S$, then multiplying on the left by $s^{-1}$ and on the right by $s$ we get $H=s^{-1}Hs$ for every $s\in S$ as well. Now let $g\in G$ be an arbitrary element. Then $g$ can be written as a product of elements of $S$ and its inverses, $g = s_1^{\epsilon_1}\cdots s_n^{\epsilon_n}$, where $\epsilon_i\in\{1,-1\}$ for every $i$; we prove that $gHg^{-1}=H$ by induction on $n$. If $n=1$, then the result holds by assumption or by our observation. If the result holds for elements that can be expressed as the product of $k$ elements of $S$ and their inverses, and $g=s_1^{\epsilon_1}\cdots s_{k+1}^{\epsilon_{k+1}}$; then by the induction hypothesis we have $$s_2^{\epsilon_2}\cdots s_{k+1}^{\epsilon_{k+1}}H(s_2^{\epsilon_2}\cdots s_{k+1}^{\epsilon_{k+1}})^{-1} = H.$$ Multiplying on the left by $s_1^{\epsilon_1}$ and on the right by $(s_1^{\epsilon_1}){-1}$, we get $$gHg^{-1} = s_1^{\epsilon_1}s_2^{\epsilon_2}\cdots s_{k+1}^{\epsilon_{k+1}}H(s_2^{\epsilon_2}\cdots s_{k+1}^{\epsilon_{k+1}})^{-1}(s_1^{\epsilon_1})^{-1} = s_1^{\epsilon_1}H(s_1^{\epsilon_1})^{-1}.$$ But by 4, $s_1^{\epsilon_1}H(s_1^{\epsilon_1})^{-1}=H$, so $gHg^{-1}=H$, as desired.

Finally, 4 implies 5 via our observation above about inverses of elements of $S$; and if 5 holds and $s\in S$, then $sHs^{-1}\subseteq H$; to show $H\subseteq sHs^{-1}$, we note that $s^{-1}Hs\subseteq H$, and multiplying on the left by $s$ and on the right by $s^{-1}$ we obtain the desired inclusion.

If $H$ is finite and 5 holds for elements $s\in S$, then since $sHs^{-1}\subseteq H$ but both are of the same size (since $x\mapsto sxs^{-1}$ is one-to-one), then they are equal, so we obtain 4 directly. $\Box$

So it suffices to check the condition for a generating set of $G$, instead of for every element of $G$. Moreover, since $hHh^{-1}=H$ for every $h\in H$, it sufffices to check the condition for every element of a generating set that is not in $H$.

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$hGh^{-1}$ contains $H$ if $H$ is a subgroup of $G$.

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Oh yes, i missed that point. in $hGh^{-1}$, $G$ is actually $G \setminus H$. Otherwise, it would not make sense as you have pointed. – foobar Mar 5 '12 at 22:40
@foobar: It's still wrong in that case; you don't get normality out of this condition. – Arturo Magidin Mar 6 '12 at 4:08

You let $g \in G$. Then show that $gHg^{-1} = H$. Since $g$ was arbitrarily chosen, $H$ is normal in $G$.

You do not need to run through all elements in $G$.

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This reminds me of the old joke: to prove that every positive integer is prime, let $n$ be an arbitrary positive integer. 17 is pretty arbitrary; and it is prime. Therefore, every positive integer is prime. – Arturo Magidin Mar 6 '12 at 4:25
When I was tutoring an Abstract Algebra class for CS department many years ago, I ran into quite a few students who believed that they need to check every element in a set if there is a $\forall$ in a problem. – scaaahu Mar 6 '12 at 4:35
My point, I guess, is that the phrasing "arbitrarily chosen" is somewhat ambiguous. What you really mean, of course, is that if one can prove that $gHg^{-1}=H$ using only the fact that $g$ is an element of $G$ and not using any other property that some elements of $G$ may have but others lack, then the argument suffices to establish the result for all elements of $G$. Much like proving $A\subseteq B$ by showing that if $a\in A$, then $a\in B$: if all you use about $a$ is that it is in $A$, you're fine. – Arturo Magidin Mar 6 '12 at 4:38
@Arturo, your explanation is much more clearer than mine. I put my answer there because I was afraid OP might just have a simple misunderstanding. He was asking a question looking like from the first chapter of an Abstract Algebra textbook. – scaaahu Mar 6 '12 at 4:48