I'm stuck to prove that the singular cohomology groups with coefficients in $\mathbb{Z}$, $H^*(S^1\times S^3;\mathbb{Z})$ and $H^*(S^1\vee S^3 \vee S^4;\mathbb{Z})$, are not isomorphic as rings.
What I've done so far: First I calculated the cohomology groups:
$H^n(S^1\vee S^3 \vee S^4)\cong H^n(S^1\times S^3)\cong\begin{cases} \mathbb{Z}, & \text{if }n=0,1,3,4\\ 0, & \text{else. } \end{cases}$
I used that $H^n(S^1\times X)\cong H^n(X)\oplus H^{n-1}(X)$, the isomorphismus $$f:H^n(X)\oplus H^{n-1}(X)\to H^n(S^1\times X)$$is given by $$(\eta_1,\eta_2)\mapsto (\pi_X^*(\eta_1)+\xi \times \eta_2),$$ with $\xi\in H^1(S^1)$ a generator and $\pi_X:S^1\times X\to X$ is the projection. Furthermore, I used that for the reduced singular cohomology it is : $H_{red}^n(S^1\vee S^3 \vee S^4)\cong H_{red}^n(S^1)\oplus H_{red}^n(S^3)\oplus H_{red}^n(S^4)$ for all $n\ge 0$.
Then we defined the cup-product in singular cohomology $$\cup: H^p(X,A;R)\otimes H^q(X,B;R)\to H^{p+q}(X,A\cup B;R)$$ by $$\cup([\alpha],[\beta]):=[\alpha\cup\beta].$$ See here cup-product for singular homology? for the definition.
The cup-product is the ring-multiplication of the cohomology groups, in our cases is $A=B=\emptyset$. Next I considered
$\cup: H^1(S^1\times S^3)\otimes H^3(S^1\times S^3)\rightarrow H^4(S^1\times S^3)$ and we can identify: $H^1(S^1\times S^3)\cong H^0(S^3)$, $H^3(S^1\times S^3)\cong H^3(S^3)$, $H^4(S^1\times S^3)\cong H^3(S^3)$.
$\cup: H^1(S^1\vee S^3 \vee S^4)\otimes H^3(S^1\vee S^3 \vee S^4)\rightarrow H^4(S^1\vee S^3 \vee S^4)$ and we can identify: $H^1(S^1\vee S^3 \vee S^4)\cong H^1(S^1)$, $H^3(S^1\vee S^3 \vee S^4)\cong H^3(S^3)$ and $H^4(S^1\vee S^3 \vee S^4)\cong H^4(S^4)$.
But now I'm stuck. I think I have to choose generators of the cohomology groups, multiply them and find out what the difference in both rings is. But I didn't know how to do this exactly, how to choose generators an so on (because I don't know, what the generators are). Could you help me to complete the proof? Best, mrs.mop
Edit: Meanwhile I considered the cup product on $H^*(S^n)$ and for the generators $1\in H^0(S^n)$, $x\in H^n(S^n)$ it is $1\cup x=1$, $x\cup 1=x$ and $x\cup x=0$. Therefore the product of the two nonzero generators 1 and x under $\cup: H^1(S^1\times S^3)\otimes H^3(S^1\times S^3)\rightarrow H^4(S^1\times S^3)$ is nonzero.
If you consider $\cup: H^1(S^1\vee S^3 \vee S^4)\otimes H^3(S^1\vee S^3 \vee S^4)\rightarrow H^4(S^1\vee S^3 \vee S^4)$, in the answer is claimed that the product of two generators is zero (I don't know how to prove this).
Can I conclude that the cohomologys are not isomorphic as rings?