As shown above, 𝑒 ∈𝑍(𝐺). Additionally, for any 𝑎,𝑏 ∈𝑍(𝐺) it is clear that 𝑎𝑏 ∈𝑍(𝐺) since 𝑎𝑏𝑥 =𝑏𝑥𝑎 =𝑥𝑎𝑏, so 𝑍(𝐺) is closed under the binary operation. Since 𝑎𝑥 =𝑥𝑎 for any 𝑥 ∈𝐺 we can both pre- and postmultiply both sides to obtain 𝑥𝑎−1 =𝑎−1𝑥 for any 𝑥, therefore 𝑎−1 ∈𝑍(𝐺), so 𝑍(𝐺) is closed under the inverse. Hence 𝑍(𝐺) is a subgroup of 𝐺 by Two step subgroup test. Now let 𝑔 ∈𝐺 and ℎ ∈𝑍(𝐺). Clearly 𝑔ℎ𝑔−1 =𝑔𝑔−1ℎ =ℎ ∈𝑍(𝐺). Hence 𝑍(𝐺) is a Normal subgroup.