Solution. The hypothesis says every element is its own inverse: \(g^{2}=e\) gives \(g^{-1}=g\). For any \(a,b\in G\) the product \(ab\) is also an element, so \[ab = (ab)^{-1} = b^{-1}a^{-1} = ba ,\] using the reversal rule for inverses and then the hypothesis on \(a\) and \(b\) separately. Hence \(G\) is abelian.

Such a group is a vector space over the field of two elements, so its order must be a power of \(2\) — which the argument above does not show, but which explains why no group of order \(6\) can have this property.

Solution. Let \([G:H]=2\), so there are exactly two left cosets, \(H\) and \(G\setminus H\), and likewise two right cosets, \(H\) and \(G\setminus H\). Take \(g\in G\).

If \(g\in H\) then \(gH = H = Hg\). If \(g\notin H\) then \(gH\) is the coset other than \(H\), so \(gH = G\setminus H\); and \(Hg\) is likewise the right coset other than \(H\), so \(Hg = G\setminus H\). In both cases \(gH = Hg\), which is normality.

The argument uses nothing about \(G\) but the count. It fails at index \(3\): the subgroup \(\left \{e,(12)\right \}\) of \(S_3\) has index \(3\) and is not normal, since \((13)(12)(13)^{-1} = (23)\).

Solution.

Normality

\(\ker \phi \) is a subgroup: \(\phi (e)=e\), and if \(\phi (a)=\phi (b)=e\) then \(\phi (ab^{-1}) = \phi (a)\phi (b)^{-1} = e\). For normality take \(k\in \ker \phi \) and \(g\in G\): \[\phi \left (gkg^{-1}\right ) = \phi (g)\phi (k)\phi (g)^{-1} = \phi (g)\,e\,\phi (g)^{-1} = e ,\] so \(gkg^{-1}\in \ker \phi \).

Injectivity

If \(\phi \) is injective then \(\phi (k)=e=\phi (e)\) forces \(k=e\). Conversely, if the kernel is trivial and \(\phi (a)=\phi (b)\), then \(\phi \left (ab^{-1}\right )=e\), so \(ab^{-1}=e\) and \(a=b\).

Checking one element instead of every pair is the practical value here, and it is why kernels are the standard route to injectivity.