The equivalence of Statements 1 and 2 is clear, as is the fact that Statement 2 implies Statement 3. So it suffices to show that Statement 3 implies Statement 2. Well, assume that \(aHa^{-1}\subseteq H\) for all \(a\in G\text{,}\) and let \(b\in G\text{.}\) We want to show that \(bHb^{-1}=H\text{.}\) Since Statement 3 holds, we clearly have \(bHb^{-1}\subseteq H\text{.}\) But, again using Statement 3, we also have \(b^{-1}Hb\subseteq H\text{;}\) multiplying both sides of this equation on the left by \(b\) and on the right by \(b^{-1}\text{,}\) we obtain \(H\subseteq bHb^{-1}\text{.}\) Hence, since \(bHb^{-1}\subseteq H\) and \(H\subseteq bHb^{-1}\text{,}\) those two sets are equal. Thus, Statement 2 is proven.

Clearly, \(e\in Z(G)\text{.}\) Next, let \(z,w\in Z(G)\text{.}\) Then for all \(a \in G\text{,}\)

so \(zw\in Z(G)\text{.}\) Finally, \(az=za\) so, multiplying both sides on the left and right by \(z^{-1}\text{,}\) we have \(z^{-1}a=az^{-1}\text{;}\) thus, \(z^{-1}\in Z(G)\text{.}\) Hence, \(Z(G)\leq G\text{.}\)

Let \(a\in G\text{.}\) Then since every element of \(Z(G)\) commutes with every element of \(G\text{,}\) every element of \(H\) commutes with every element of \(G\text{;}\) so we must have \(aH=Ha\text{.}\) Thus, \(H\unlhd G\text{.}\)

Let \(K=\Ker \phi\text{.}\) We first show that \(K\) is a subgroup of \(G\text{.}\) Clearly, the identity element of \(G\) is in \(K\text{,}\) so \(K\neq \emptyset\text{.}\) Next, let \(k,m\in K\text{.}\) Then, letting \(e'\) denote the identity element of \(G'\text{,}\) we have

so \(km^{-1}\in K\text{.}\) Thus, by the Two-Step Subgroup Test, we have that \(K\) is a subgroup of \(G\text{.}\)

Next, let \(k\in K\) and let \(a\in G\text{.}\) Then

So \(aka^{-1}\in K\text{.}\) Thus, \(K \unlhd G\text{.}\)