I seem to be twisting myself into pretzels,can someone help me with the following question?
Let $(G,*)$ be a group and $H\unlhd G$ and $K\unlhd G$ (i.e $H$ and $G$ are normal subgroups of $G$).Prove that $HK$ is a subgroup of $G$,where $HK=\{hk:h \in H \text{ and } k \in K\}$.
The book gives the [hint] to use the following :
Let $(G,*)$ be a group and let $a \in G$. Suppose $N\unlhd G$ then the following holds:
- For all $n \in N$ there exists a $j \in N$ such that na=aj.
- For all $n \in N$ there exists a $k \in N$ such that an=ka.
Since different books may follow different approaches,the lemma that seems to be used throughout mine is the following:
Lemma(Subgroup Lemma)
Let $(G,*)$ be a group and $H \subseteq G$. Then $H$ is a subgroup of $G$ if and only if the following holds:
- (a) $e \in H$ (where e is the identity element of G)
- (b) $ab \in H$ whenever $a,b \in H$
- (c) $a^{-1} \in H$ whenever $a \in H$
My twisted attempt:
Let $(G,*)$ be a group and $H\unlhd G$ and $K\unlhd G$ (i.e $H$ and $G$ are normal subgroups of $G$).Suppose $HK=\{hk:h \in H \text{ and } k \in K\}$. To prove that $HK$ is a subgroup of $G$,the following must hold:
(i)$HK \subseteq G$ , (ii) Subgroup Lemmas defined proven true.
Regarding (i)
Let $h \in H$ and $k \in K$.With $H,K \subseteq G$, then $h,k \in G$. Since $(G,*)$ is a group then$(h*k) \in G$ and $HK \subseteq G$ as required ?
Regarding (ii)
(a) Prove $e \in HK$
Since $(G,*)$ is a group then it satisfies the three group axioms and there exists an element $e \in G$ such that for all $g\in G$, $g*e=e*g=g$. Since $H$ and $K$ are normal subgroups of $G$ then by subgroup lemma, $e \in H$ and $e \in K$. With $(e*e) \in HK$ and $(e*e)=e$ then $e \in HK$ as required ?
(b) Prove $ab \in HK$ whenever $a,b \in HK$
Let $a,b \in HK$ , then by definition of $HK$ , $a=hk$ and $b=ij$ for some $h,i \in H$ and $k,j \in K$. the objective is to prove that $(ab) \in HK$, or equivalently that $(ab)=cd$ for some $c \in H$ and $d \in K$.
Let $g \in G$ and note that by [hint above] : since $i \in H$ there exists some $r \in H$ such that ig=gr and therefore ij=jr. With $(ab)=(hk)(ij)=(hk)(jr)=h(kj)r$ by the associative axiom of groups.
Let d=(kj) and note that since $k,j \in K$ and $K \unlhd G$ then $(kj) \in K$ and therefore $d \in K$.
With $(ab)=h(kj)r=hdr$,the [hint above] suggests that : since $d \in K$, there exists a $p \in K$ such that $dg=gp$ and therefore $dr=rd$. Hence $(ab)=h(kj)r=h(dr)=h(rd)=(hr)d$.
Letting $c=(hr)$ then $c \in H$ since $h,r \in H$ and $H \unlhd G$ imply that $(hr) \in H$.
Finally $(ab)=cd$ as required?
(c) Prove $a^{-1} \in H$ whenever $a \in H$
...will do later if the attempt so far is OK
Thank you for your help
The ideas in your proof, in broad lines, seem correct to me (although there would be a faster way to prove the statement, i.e. using the lemma just once).
It seems to me, anyway, that you are using a stronger (and wrong) version of the lemma: when you say, in the second paragraph of (b), that "there exists some $r\in H$ such that ig=gr and therefore ij=jr", I think you are making a mistake. The lemma does not assert that there exists a $r$ that does the job for every $g\in G$, it just tells you that for every $g$ such a $r_g$ can be found. This is exactly what you need for the proof anyway, so with this caveat your proof works.