If $G$ is a group and $K$ is a subgroup, is $gKg^{-1}$ a subgroup? My lecturer stated that If $G$ is a group and $K$ is a subgroup then $gKg^{-1}$ is a subgroup also, whilst proving the sylows theorem, I'm wondering how does this hold?
 A: Let 
$$\varphi: K \to G,\; k\mapsto gkg^{-1}$$
then
$$\varphi(kk')=gkg^{-1}gk'g^{-1}=\varphi(k)\varphi(k')$$
which means that $\varphi$ is a homomorphism of groups so $gKg^{-1}=\varphi(K)$ is a subgroup of $G$.
A: Let $gk_1g^{-1}, gk_2g^{-1} \in gKg^{-1}.$ Then $gk_1g^{-1} (gk_2g^{-1})^{-1}=gk_1k_2^{-1}g^{-1} \in gKg^{-1}.$
A: Just check the conditions for being a subgroup one by one. 


*

*Is $1_G \in gKg^{-1}$ true? Yes as $1_G \in K$ and $g1_Gg^{-1} = 1_G$.

*For $h,h' \in gKg^{-1} $ do you have $hh' \in gKg^{-1}$? Yes, as $h = gkg^{-1}$ and $h' = gk'g^{-1}$ for some $k, k' \in K$ and  so $hh' = g (kk')g^{-1} \in gKg^{-1}$

*For $h \in gKg^{-1}$ do you have $h^{-1} \in gKg^{-1}$? Yes, as $h = gkg^{-1}$ and  $   hgk^{-1}g^{-1}= 1_G$. 
There are shorter arguments using the criterion that a non-empty set is a subgroup if and only if it contains $h_1h_2^{-1}$, or also by showing that conjugation is an isomorphism and isomorphisms preserved subgroups (as given in another answer). 
A: 
Theorem: Let $H$ be a nonempty subset of a group $G$. Then $H$ is a subgroup
  of $G$ if and only if $ab^{−1} \in H$ for all $a, b \in H$.

The proof is in every book of Algebra. (Algebra of Thomas W. Hungerford
for example)
Then, let $gkg^{−1}$ and $ghg^{−1}$ with $k, h \in K$ two elements of $gKg^{−1}$. We only
have to see that, $gkg^{−1} \cdot (ghg^{−1} )^{−1} \in gKg^{−1}$:
$gkg^{−1} \cdot (ghg^{−1} )^{−1} = gkg^{−1} \cdot gh^{−1} g^{ −1} = gkh^{−1} g^{−1} \in gKg^{−1}$ ( $gh^{−1} \in K$
because $K$ is a subgroup).
