Let $G$ be a group of order $15$ and $A$ and $B$ subgroups of $G$ of order $5$ and $3$, respectively. Show that $G= AB$

Question

Let $G$ be a group of order $15$ and $A$ and $B$ subgroups of $G$ of order $5$ and $3$, respectively. Show that $G= AB.$

I tried solving the problem:

Since $o(A),o(B)$ are prime, so, $A,B$ are cyclic. Also, $o(A)\neq o(B)$ and thus they are unique, due to which $A\cap B=\{e\}.$ Also, $|AB|=15=|G|.$ Now, $G$ is cyclic and $A,B$ are normal subgroups.

But I cant proceed further as there's no gurantee that there will be no repitition in $\{ab,\forall a\in A,B\in B\}$ ? I don't know, how to proceed further,i.e the short question how to prove $A,B$ as internal direct product of $G$?

Answer 1

There are certainly tons of duplicates on this site but I have not been able to find any. I stay ready to delete my answer if someone is more successful.

For any two finite subgroups $A,B$ of a group $G,$ $|AB|= \frac{|A||B|}{|A\cap B|}.$

If moreover $m:=|A|$ and $n:=|B|$ are coprime (like in your example) then (by Lagrange) their common subgroup $A\cap B$ is trivial hence $|AB|=mn.$

Therefore, if additionnaly $|G|=mn,$ then $AB=G.$

Answer 2

It is possible to use Schur's Theorem: Let A be an abelian normal subgroup of a finite group G. Let's suppose that $mdc(|A|,[G:A])= 1$, then there exists a complement for A in G, that is , there exists a subgroup K of G such that G = AK and $A \cap K =1$. For the case it is known that G is cyclic and therefore abelian, so its subgroups are abelian and normal in G. Taking any non-trivial subgroup A we see that $mdc(|A|,[G:A])= 1$. The result follows.

Answer 3

Anne's solution is quite optimal. They liked and accepted it. I wonder how my colleague Nuray would solve this question. Because we have all different paths of thinking and experiences.

My try: Groups of prime order are cyclic. So, $A=<x>$ and $B=<y>$ where $o(x)=5$ and $o(y)=3$ respectively. Note: $o(g)$ is the notation for the order of the element $g$.

Now, I claim that $$AB=\{e,x,x^2,x^3,x^4,y,y^2,xy,xy^2,x^2y,x^2y^2,x^3y,x^3y^2,x^4y,x^4y^2\}$$ and $|AB|=15$. Proof: Suppose that $x^iy^j=x^ky^l$ and these products are in $AB$ (symbolically). Then, $x^{i-k}=y^{l-j}$. Taking third power, we have $x^{3(i-k)}=e$ and $5|i-k$ which implies that $i=k$. Similarly, taking fifth power, we get $j=l.$

Now, we have that $AB\subset G$ and $|AB|=|G|$. This implies that $AB=G.$ It turns out that $AB$ is a group. Note that this is not always the case. We must also note that $G$ is cyclic becuse there is only one group of order $15$. Also, I didn't use the technical product formula of Anne in my solution. If we do not know the proofs of formulas well we hesitate to use them.