Question
Let $G$ be a group with $|G|=595=5 \cdot 7 \cdot 17$. Show that the Sylow 5-subgroup of $G$ is normal in $G$ and is contained in $Z(G)$.
Step 1
By Sylow's Theorem, the number of Sylow 5-subgroups in G, denoted by n_5, must divide the order of G and be congruent to 1 modulo 5. Thus, n_5 must divide 7 * 17 = 119. The only divisors of 119 that are congruent to 1 modulo 5 are 1 and 119. If n_5 = 119, then Show more…
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