# Show that $\{ I,(12)(34),(13)(24),(14)(23)\}$ subgroup of $A_4$ is invariant

Let $$H:= (h_1,h_2,...)$$ be a subgroup of $$G$$. Take any element g, not in $$H$$. Then the set of elements $$(g^{-1}h_1 g,g^{-1}h_2 g,...)$$ forms a subgroup, which we denote by $$g^{-1}Hg$$. If H and $$g^{-1}Hg$$ are the same for all $$g \in G$$, then $$H$$ is invariant.

How can I show that a subgroup of $$A_4$$,$$\{ I,(12)(34),(13)(24),(14)(23)\}$$ forms an invariant group and $$\{I,(12)(34),(13)(24),(14)(23) \}\cong\Bbb Z_2 \otimes\Bbb Z_2$$

Reference: Group Theory in a Nutshell for Physicists, A.Zee

I tried several other $$A_4$$ elements such as $$(123), (142),$$ but I somehow cannot write $$(142)^{-1}(12)(34)(142)$$ into another element of the subgroup above...

Since the conjugation $$\beta^{-1}\alpha\beta$$ of a permutation $$\alpha$$ by another permutation $$\beta$$ preserves the cycle structure of $$\alpha$$, and the candidate subgroup $$H$$ in question contains all elements of the form $$(ab)(cd)$$ (together with the identity), we must have that $$H\lhd A_4$$, which is equivalent to $$H$$ being invariant.
Since $$H$$ is closed under multiplication, $$H< A_4$$. Thus $$H$$ is, in particular, a group. Each of its nontrivial elements has order two; therefore, by the classification of groups of order four, $$H\cong \Bbb Z_2^2$$.