I am faced with the following problem:
(a) Prove that for any groups $G_{1}$ and $G_{2}$, $|Aut(G_{1}\times G_{2})| \geq |Aut(G_{1})| \cdot |Aut(G_{2})|$.
(b) Is it true that for any $G_{1}$ and $G_{2}$ we have the equality $|Aut(G_{1} \times G_{2})| = |Aut(G_{1})| \cdot |Aut(G_{2})|$?
I have a solution for (a):
Suppose $\alpha \in Aut(G_{1})$, $\beta \in Aut(G_{2})$. By definition, $\alpha$ is a bijective homomorphism from $G_{1}$ into $G_{1}$ and $\beta$ is a bijective homomorphism from $G_{2}$ into $G_{2}$. Now, consider the product $\alpha \times \beta$ and an element $(x,y) \in G_{1} \times G_{2}$. Then, $(\alpha \times \beta)(x,y) = (\alpha(x), \beta(y))$.
First, we show that $(\alpha \times \beta)$ is a homomorphism:
Consider $(x_{G_{1}}, x_{G_{2}})$, $(y_{G_{1}}, y_{G_{2}}) \in G_{1} \times G_{2}$. Then $(\alpha \times \beta)\left[(x_{G_{1}},x_{G_{2}})\times (y_{G_{1}}, y_{G_{2}}) \right] = (\alpha \times \beta)(x_{G_{1}}y_{G_{1}}, x_{G_{2}}y_{G_{2}})= (\alpha(x_{G_{1}}y_{G_{1}}), \beta(x_{G_{2}},y_{G_{2}})) = \left(\alpha(x_{G_{1}})\alpha(y_{G_{1}}), \beta(x_{G_{1}})\beta(y_{G_{1}})\right)\,\text{(Since }\,\alpha\, \text{and}\, \beta\, \text{are each homomorphisms)} = (\alpha(x_{G_{1}}),\beta(x_{G_{2}})) \times (\alpha(y_{G_{1}}),\beta(y_{G_{2}})) = (\alpha \times \beta)(x_{G_{1}},x_{G_{2}}) \times (\alpha \times \beta)(y_{G_{1}},y_{G_{2}})$. So, $(\alpha \times \beta)$ is a homomorphism.
Next, we must show that $(\alpha \times \beta) $ is injective:
Suppose that $\forall (x,y) \in G_{1} \times G_{2}$, $(\alpha_{1} \times \beta_{1})(x,y) = (\alpha_{2} \times \beta_{2})(x,y)$ $\implies$ $(\alpha_{1}(x), \beta_{1}(y)) = (\alpha_{2}(x), \beta_{2}(y)) $.
Now, we must show that $\alpha_{1} = \alpha_{2}$.
Suppose that it is not, then $\exists$ some $x^{\prime} \in G_{1}$ such that $\alpha_{1}(x^{\prime}) \neq \alpha_{2}(x^{\prime})$.
Then, consider $(\alpha_{1} \times \beta_{1})(x^{\prime}, e_{G_{2}}) = (\alpha_{2} \times \beta_{2})(x^{\prime}, e_{G_{2}})$ $\implies $ $(\alpha_{1}(x^{\prime}), \beta_{1}(e_{G_{2}})) = (\alpha_{2}(x^{\prime}),\beta_{2}(e_{G_{2}}))$.
But, since $\beta_{1}$ and $\beta_{2}$ are injective, $\beta_{1}(e_{G_{2}}) = e_{G_{2}}$ and $\beta_{2}(e_{G_{2}}) = e_{G_{2}}$.
So, we have that $(\alpha_{1} \times \beta_{1})(x^{\prime}, e_{G_{2}}) = (\alpha_{2} \times \beta_{2}) (x^{\prime}, e_{G_{2}}) \, \implies \, (\alpha_{1}(x^{\prime}), e_{G_{2}}) = (\alpha_{2}(x^{\prime}), e_{G_{2}}) \, \implies \, \alpha_{1}(x^{\prime}) = \alpha_{2}(x^{\prime})$ when we assumed it was not (contradiction).
Then, I did something very similar for $\beta$ to show that $\forall y^{\prime} \in G_{2}$, we must have that $\beta_{1}(y^{\prime})=\beta_{2}(y^{\prime})$.
Together, these things tell us that $\alpha \times \beta$ is an injective homomorphism, and since it essentially maps the images of $x \in G_{1}$ under $\alpha$ and the images of $y \in G_{2}$ under $\beta$ to the images in $G_{1} \times G_{2}$ under $\alpha \times \beta$, its being injective tells us that $|Aut(G_{1} \times G_{2})| \geq |Aut(G_{1})| \times |Aut(G_{2})|$.
Now, my question is how to prove part (b).
I'm assuming that in order to show that $|Aut(G_{1} \times Aut(G_{2})| = |Aut(G_{1})| \cdot |Aut(G_{2})|$, we must, in addition to showing that $|Aut(G_{1} \times G_{2})| \geq |Aut(G_{1})| \times |Aut(G_{2})|$, we must show the reverse inequality: $|Aut(G_{1} \times G_{2})| \leq |Aut(G_{1})| \times |Aut(G_{2})|$.
Since we used injectivity to prove the $\geq$ direction, I'm assuming we would use surjectivity to prove the $\leq$ direction?
However, I am inclined to believe that the answer is false. I have heard that for finite groups $G_{1}$ and $G_{2}$, it doesn't work unless $\gcd(|G_{1}|, |G_{2}|) = 1$, so I'd want to state that (b) is not true except for when $|G_{1}|$ and $|G_{2}|$ are finite and $\gcd(|G_{1}|, |G_{2}|) = 1$. But, 1) I'm not sure how to prove this, and 2) would I then effectively be proving that this is the ONLY case where we have the $\leq$ direction (I sometimes tend to overthink the logic; be patient with me, I'm learning!)?
If I could get some written out help for these things, I would be most appreciative!
Consider the example where $G_1=G_2=Z/2Z$ and compute the automorphism groups.