Let \(G\) be a group of order \(120\) and \(H\le G\) a subgroup of order \(24\). Assume that at least one left coset of \(H\), other than \(H\) itself, is equal to some right coset of \(H\). Prove that \(H\) is normal in \(G\).
Solution
Let \(gH\) be a left coset with \(g\notin H\) that is equal to a right coset \(Hx\).
1. One has\[ gH=Hg. \]
Since \[ g\in gH=Hx, \] there is some \(h\in H\) such that \(g=hx\). Hence \[ x=h^{-1}g, \] and therefore \[ Hx=Hh^{-1}g=Hg. \] Thus the assumed equality \(gH=Hx\) becomes \(gH=Hg\).
2. The element \(g\) lies in the normalizer \(N_G(H)\), and\(g\notin H\).
By step 1, \[ gH=Hg. \] Multiplying on the right by \(g^{-1}\) gives \[ gHg^{-1}=H, \] so \(g\in N_G(H)\). The chosen left coset was not \(H\) itself, so \(g\notin H\).
3. The normalizer satisfies\[ H<N_G(H)\le G. \]
Every subgroup normalizes itself, so \(H\le N_G(H)\). Step 2 gives an element of \(N_G(H)\) outside \(H\), making the first inclusion strict. The second inclusion is part of the definition of the normalizer.
4. One has\[ N_G(H)=G. \]
The index of \(H\) in \(G\) is \[ [G:H] = \frac{120}{24} = 5. \] By multiplicativity of indices along the subgroup chain in step 3, \[ 5 = [G:N_G(H)]\,[N_G(H):H]. \] Since \(H<N_G(H)\), the factor \([N_G(H):H]\) is greater than \(1\). Because \(5\) is prime, it must equal \(5\), and therefore \([G:N_G(H)]=1\). Hence \(N_G(H)=G\).
5. Consequently,\[ \boxed{H\trianglelefteq G}. \]
Step 4 says that every element of \(G\) normalizes \(H\), which is exactly the definition that \(H\) is normal in \(G\).
6. Q.E.D.
Step 5 is the required conclusion.