# How is this subgroup normal?

Let $G$ be a group, and let $U$ be a subset of $G$. Let $\hat{U}$ be the smallest subgroup of $G$ containing $U$. Then $\hat{U}$ is the intersection of the collection of all the subgroups of $G$ containing $U$. (Right?) This collection is obviously non-empty as $G$ itself is a subgroup of itself which contains $U$.

Now my question is this:

If $gug^{-1} \in U$ for all $g \in G$ and $u \in U$, then is the subgroup $\hat{U}$ a normal subgroup of $G$?

If the answer is yes, then one way of proving this is to find a homomorphism $\phi \colon G \to G$ with kernel equal to $\hat{U}$.

If the answer is no, then what counter-example can we give?

• Nevermind - I did misunderstand Apr 14, 2014 at 18:01
• @FireGarden But then the condition on $U$ that $gug^{-1} \in U$ for all $u \in U$ and $g \in G$ would fail. Apr 14, 2014 at 18:02
• user2566092, how would this condition fail? I just didn't get your comment. Apr 14, 2014 at 18:22
• Can't anyone give me a homomorphism for which $\hat{U}$ is a kernel? Apr 15, 2014 at 1:58

The condition $gug^{-1} \in U$ for all $g\in G$ and $u\in U$ means that we have $gUg^{-1} = U$ for all $g\in G$. For if we had $gUg^{-1} \subsetneq U$ for some $g$, let $D = U \setminus gUg^{-1}$, then

$$g^{-1}Ug = g^{-1}\left( gUg^{-1} \cup D\right)g = g^{-1}gUg^{-1}g \cup g^{-1}Dg = U \cup g^{-1}Dg,$$

and $g^{-1}Dg\neq\varnothing$, but $U\cap g^{-1}Dg = \varnothing$, since the conjugation $x\mapsto g^{-1}xg$ is a bijection.

Thus, for any subgroup $H\subset G$, we have

$$U\subset H \iff g^{-1}Ug\subset H \iff U \subset gHg^{-1}$$

for all $g\in G$. In particular, since $U\subset \hat{U}$, we have $U \subset g\hat{U}g^{-1}$ for all $g\in G$, and since $\hat{U}$ is the smallest subgroup containing $U$,

$$\hat{U} \subset g\hat{U}g^{-1}$$

for all $g\in G$, which is another way to characterise the normality of a subgroup. Thus indeed, $\hat{U}$ is normal.

There's an explicit characterization of the smallest subgroup containing a given set: $$\hat U=\{u_1^{\varepsilon_1}\ldots u_n^{\varepsilon_n}\mid u_i\in U,\varepsilon_i\in\{-1,1\}\text{ and }n\ge0\text{ (not a mistake)}\}$$ Because it must contain all these elements and it's closed under all group operations (including the identity).

Now observe that for any $g\in G$ we have $$gu_1^{\varepsilon_1}\ldots u_n^{\varepsilon_n}g^{-1}=gu_1^{\varepsilon_1}g^{-1}\ldots gu_n^{\varepsilon_n}g^{-1}=(gu_1g^{-1})^{\varepsilon_1}\ldots(gu_ng^{-1})^{\varepsilon_n}={u'_1}^{\varepsilon_1}\ldots{u'_n}^{\varepsilon_n}$$ So $\hat U$ is a normal subgroup of $G$.

• user2345215, can you please also define a homomorphism from $G$ into itself for which $\hat{U}$ be the kernel? Apr 16, 2014 at 13:51
• @SaaqibMahmuud I don't think that can be done. If $\varphi:G\to G$ and $\ker\varphi=\hat U$, then $G/\hat U\simeq\varphi(G)$, so we have an embedding of $G/\hat U$ inside $G$. This is not possible e.g. for $\mathbb Z$ and a subgroup $2\mathbb Z$, because $\mathbb Z/2\mathbb Z\simeq\mathbb Z_2$ which is not a subgroup of $\mathbb Z$ Apr 16, 2014 at 14:23

We may as well assume $e \in U$ because $e \in \hat{U}$. Then the closure of $U$ contains the set of all products $ab^{-1}$ for all $a,b \in U$. Then maybe this construction has to be repeated on the set of products obtained (I'm not sure), but clearly if the construction is repeated indefinitely and you take the union then you will have the subgroup $\hat{U}$. So all we have to do is show that if $a,b \in U$ then $g (ab^{-1}) g^{-1} \in \hat{U}$ for all $g \in G$. But note that $g (ab^{-1}) g^{-1} = (gag^{-1})(gbg^{-1})^{-1}$, which is a product of the form $a'b'^{-1}$ where $a',b' \in U$ and thus is contained in $\hat{U}$. So $\hat{U}$ is normal.

This is true. Let's call a subset of $G$ normal iff it is closed under conjugation. Let $T=U\cup\{u^{-1}|u\in U\}$. Let $g\in G$ be any element. If $t\in T$, then either $t\in U$, therefore $gtg^{-1}\in U$ or $t=u^{-1}$ for some $u\in U$, therefore $gtg^{-1}=gu^{-1}g^{-1}=(gug)^{-1}$ since $gug^{-1}\in U$, then $t=(gug)^{-1}\in T$. This shows that $T$ is normal. Now clearly, $\overline{U}=\overline{T}$. Let $x\in \overline{T}$ , then $x=t_1t_2...t_n$ for some $t_1,...,t_n\in T$. For any $g\in G$, we have:

$$gxg^{-1}=g(t_1t_2...t_n)g^{-1}=(gt_1g^{-1})(gt_2g^{-1})...(gt_ng^{-1})$$

Which shows that $\overline{U}=\overline{T}\lhd G$, because $gt_1g^{-1},gt_2g^{-1},...,gt_ng^{-1}\in T$

• Amr, what is meant by $\overline{T}$ etc? Apr 15, 2014 at 1:32
• @SaaqibMahmuud The group generated by $T$
– Amr
Apr 15, 2014 at 9:18