Let’s carefully work through the Zassenhaus (Butterfly) Lemma proof step by step, using the Fundamental Theorem of Homomorphism (First Isomorphism Theorem) for groups. --- ## Step 1: Restating the given setup We have: - \( G \) is a group. - \( H, K \) are subgroups of \( G \). - \( H' \trianglelefteq H \) (\( H' \) normal in \( H \)), \( K' \trianglelefteq K \). We want to prove: 1. \( H'(H \cap K') \trianglelefteq H'(H \cap K) \) 2. \( K'(H' \cap K) \trianglelefteq K'(H \cap K) \) 3. \[ \frac{H'(H \cap K)}{H'(H \cap K')} \simeq \frac{K'(H \cap K)}{K'(H' \cap K)} \simeq \frac{H \cap K}{(H' \cap K)(H \cap K')}. \] The notation \( H'(H \cap K) \) means the product of subgroups \( H' \) and \( H \cap K \) inside \( G \). Since \( H' \trianglelefteq H \), \( H'(H \cap K) \) is a subgroup of \( H \). --- ## Step 2: Prove \( H'(H \cap K') \trianglelefteq H'(H \cap K) \) First, note: - \( H' \trianglelefteq H \) ⇒ \( H' \trianglelefteq H'(H \cap K) \) (since \( H' \) is normal in \( H \) and \( H'(H \cap K) \leq H \)). Also, \( H \cap K' \) is normal in something — we should check: \( K' \trianglelefteq K \) ⇒ \( H \cap K' \trianglelefteq H \cap K \) (because for any \( x \in H \cap K \), conjugation by \( x \) is inside \( K \) and preserves \( K' \) since \( K' \trianglelefteq K \), and \( x \) also in \( H \) so conjugates \( H \cap K' \) within \( H \cap K' \)). So \( H \cap K' \trianglelefteq H \cap K\). Thus \( H'(H \cap K') \) is a subgroup (product of two subgroups \( H' \) and \( H \cap K' \) with \( H' \trianglelefteq H \) and \( H \cap K' \leq H \)). Since \( H' \trianglelefteq H \) and \( H \cap K' \trianglelefteq H \cap K \), normality in \( H'(H \cap K) \) requires checking: Take \( a \in H'(H \cap K) \). Write \( a = h_1 c \) with \( h_1 \in H' \) and \( c \in H \cap K \). We want to show: \( a \cdot H'(H \cap K') \cdot a^{-1} \subseteq H'(H \cap K') \). Let \( x \in H'(H \cap K') \), write \( x = h_2 d \) with \( h_2 \in H' \), \( d \in H \cap K' \). Then \[ a x a^{-1} = (h_1 c)(h_2 d)(c^{-1} h_1^{-1}) = h_1 [c h_2 c^{-1}] [c d c^{-1}] h_1^{-1}. \] Now: \( c h_2 c^{-1} \in H' \) because \( H' \trianglelefteq H \), \( c \in H \). \( c d c^{-1} \in H \cap K' \) because \( d \in H \cap K' \trianglelefteq H \cap K \), \( c \in H \cap K \). So \( c h_2 c^{-1} \in H' \), \( c d c^{-1} \in H \cap K' \). Thus \( [c h_2 c^{-1}][c d c^{-1}] \in H'(H \cap K') \). Call this \( y \). Then \( a x a^{-1} = h_1 y h_1^{-1} \). But \( h_1 y h_1^{-1} \in H'(H \cap K') \) because \( h_1 \) normalizes \( H' \) trivially and \(