Let \(G\) be a group with order 112. We show that \(G\) is not simple.
Note that \(|G| = 2^4\cdot7\).
Let us assume, for sake of contradiction, that \(G\) is simple.
Then, \(n_2 > 1\) and \(n_7 > 1.\)
The constraints given by Sylow Theorem (3) force \(n_2 = 7.\)
Let \(G\) act on Syl\(_2(G)\) by conjugation.
By Sylow Theorem (2), we know that this is a transitive action.
Consider the corresponding natural homomorphism \(m:G\to S_7.\)
By transitivity of the action, we know that ker \(m \neq G.\) However, \(G\) is simple by assumption and kernel is a normal subgroup. Thus, ker \(m = (1).\)
This gives us that \(G\) is isomorphic to a subgroup \(H\) of \(S_7.\) (To be precise, \(H = m(G).\))
Let us now analyse this subgroup \(H\) and arrive at a contradiction.
As \(|H| > 2,\) it must have an even cycle distinct from 1. (This is easy to argue.)
This shows that \(K = H \cap A_7\) is nontrivial. (\(A_7\) is the subgroup of even cycles.)
Claim: \(K\) is normal in \(H.\)
Proof. Let \(h \in H.\) We show that \(hKh^{-1} = K.\)
Let \(x \in hKh^{-1}.\) Then, \(x = hkh^{-1}\) for some \(k \in K.\)
As \(k \in K \subset H,\) we know that \(x \in H,\) by closure under products and inverses.
Now, \(k \in A_n\) and is thus, an even cycle. Therefore, its conjugate is also even. (This can be argued by using the fact that \(A_n\) is normal or counting the number of transpositions.)
Thus, \(k \in H \cap A_n = K.\) This gives us \(hKh^{-1} \subset K.\)
Equality follows by considering the fact that \(|hKh^{-1}| = |K|. \quad \blacksquare\)
Thus, \(K\) is a normal subgroup of \(H\) different from \((1).\) This forces \(K = H,\) as \(H\) is simple.
Thus, \(H \cap A_n = H\) and therefore, \(H \subset A_n.\) Moreover, \(H \le A_n.\)
However, \(|H| = 112\) and \(|A_n| = 2520,\) which leads to a contradiction as \(112 \not\mid 2520.\)
Thus, we are done!