Question

Prove that a subgroup N of a group G is normal if and only if it has this property: $ab\in N$ if and only if $ba\in N$, for all $a,b\in G$ .

Short answer

It has been proved that a subgroup N of a group G is normal if and only if it has this property: $ab\in N$ if and only if $ba\in N$, for all $a,b\in G$.

Step-by-step solution

Suppose that N  is normal

It is given that N is a subgroup of G .

Suppose N is normal.

Let $a,b\in G$ and $ab\in N$,

then $ba=b\left(ab\right)b^{-1}\in bN{b}^{-1}$

Since $bN{b}^{-1}=N$

So, $ba\in N$

Hence the property is proved.

Similarly, we can prove that $ab\in N$if $ba\in N$.

Suppose that  N has the property

Suppose that N has the property: $ab\in N$ if and only if $ba\in N$, for all $a,b\in G$.

Let $g\in G$ and $n\in N$

We have $g\left(g^{-1}n\right)=n\in N$.

The above-mentioned property implies $g^{-1}ng\in N$

Therefore, $g^{-1}Ng\subseteq N$ for every g .

Thus, N is normal.

Conclusion

Hence, a subgroup N of a group G is normal if and only if it has this property: $ab\in N$ if and only if $ba\in N$, for all $a,b\in G$.